Zeal Education Society’s
ZEAL POLYTECHNIC, PUNE.
NARHE │PUNE -41 │ INDIA
FIRST YEAR (FY)
DIPLOMA IN COMPUTER ENGINEERING
SCHEME: I SEMESTER: II
NAME OF SUBJECT: PROGRAMMING IN C
Subject Code: 22226
MSBTE QUESTION PAPERS & MODEL ANSWERS
1. MSBTE SUMMER-18 EXAMINATION
2. MSBTE WINTER-18 EXAMINATION
3. MSBTE SUMMER-19 EXAMINATION
4. MSBTE WINTER-19 EXAMINATION
22226
[1 of 4] P.T.O.
21718
3 Hours / 70 Marks Seat No.
Instructions : (1) All Questions are compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Mobile Phone, Pager and any other Electronic Communication
devices are not permissible in Examination Hall.
(7) Preferably, write the answers in sequential order.
Marks
1. Attempt any FIVE of the following : 10
(a) Define :
(i) Two dimensional array
(ii) Multi-dimensional array
(b) Give any four advantages of pointer.
(c) Define type casting. Give any one example.
(d) State any four decision making statements.
(e) State any four math functions with its use.
(f) State the use of following symbols used for flowchart drawing :
(i)
(ii)
(iii) (iv)
(g) State use of while loop with syntax.
22226 [2 of 4]
2. Attempt any THREE of the following : 12
(a) Develop a simple ‘C’ program for addition and multiplication of two integer
numbers.
(b) Explain how to pass pointer to function with example.
(c) Explain following functions :
getchar( )
putchar( )
getch( )
putch( )
with suitable examples.
(d) Develop a program to accept an integer number and print whether it is
palindrome or not.
3. Attempt any THREE of the following : 12
(a) State the use of printf( ) & scanf( ) with suitable example.
(b) Explain any four library functions under conio.h header file.
(c) Explain how formatted input can be obtain, give suitable example.
(d) Develop a program to find factorial of a number using recursion.
22226 [3 of 4]
P.T.O.
4. Attempt any THREE of the following : 12
(a) Write a program to sweep the values of variables a = 10, b = 5 using function.
(b) Develop a program using structure to print data of three students having data
members name, class, percentage.
(c) Design a program to print a message 10 times.
(d) Draw a flowchart for checking whether given number is prime or not.
(e) Implement a program to demonstrate logical AND operator.
5. Attempt any TWO of the following : 12
(a) Draw a flowchart of Do-while loop and write a program to add numbers until
user enters zero.
(b) Give a method to create, declare and initialize structure also develop a
program to demonstrate nested structure.
(c) Implement a program to demonstrate concept of pointers to function.
6. Attempt any TWO of the following : 12
(a) Develop a program to swap two numbers using pointer and add swaped
numbers also print their addition.
(b) Design a programme in C to read the n numbers of values in an array and
display it in reverse order.
(c) Develop a program to find diameter, circumference and area of circle using
function.
_______________
22226 [4 of 4]
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Subject: Programming in ‘C’ Subject Code:
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22226
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance
(Not applicable for subject English and Communication Skills).
4) While assessing figures, examiner may give credit for principal components indicated in the
figure. The figures drawn by candidate and model answer may vary. The examiner may give
credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate‟s answers and model
answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant
answer based on candidate‟s understanding.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
Q.
No
.
Sub
Q.N.
Answer
Marking
Scheme
1.
(a)
Ans.
Attempt any FIVE of the following:
Define:
(i) Two dimensional array
(ii) Multi-dimensional array
(i) Two dimensional array
Two dimensional array is a collection of similar type of data elements
arranged in the form of rows & columns.
E.g. Array can be declared as int arr[3][3];
In this there can be 9 elements in an array with 3 rows and 3 columns.
(ii) Multi-dimensional array:
An array with more than one dimension is called as multi-
dimensional array.
For example,
float x[3][4];
Similarly, you can declare a three-dimensional (3d) array. For
example,
float y[2][4][3];
10
2M
Definitio
n of two-
dimensi
onal
array
1M
Multi-
dimensi
onal
array
1M
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22226
Here, The array y can hold 24 elements.
(b)
Ans.
Give any four advantages of pointer.
Advantages of pointer:
1. Pointers reduce the length and complexity of a program.
2. They increase execution speed.
3. A pointer enables us to access a variable that is defined
outside the function.
4. Pointers are more efficient in handling the data tables.
5. The use of a pointer array of character strings results in
saving of data storage space in memory.
6. It supports dynamic memory management.
2M
Any
four
advanta
ges ½M
each
(c)
Ans.
Define type casting. Give any one example.
Definition type casting:
The conversion of one data type to another is known as type casting.
The values are changed for the respective calculation only, not for
any permanent effect in a program.
For example,
x=int (7.5) means 7.5 is converted to integer by truncating it i.e. 7
b=(int) 22.7/(int) 5.3 means 22.7 will be converted to 22 and 5.3 to 5
so answer will be 22/5=4
c=(double) total/num means the answer will be in float value.
p=sin((int)x) means x will be converted to integer and then sine angle
will be calculated.
2M
Definitio
n of type
casting
1M
Any one
correct
Example
1M
(d)
Ans.
State any four decision making statement.
Decision making statement:
1. if statement
2. if-else statement
3. if-else-if ladder
4. Nested if-else statement
5. switch statement
6. conditional operator statement (? : operator)
2M
Any
four
correct
decision
making
statemen
ts - ½ M
each
(e)
Ans.
State any four math functions with its use.
(Note: Any other relevant math function shall be considered)
Math Functions:
sqrt() - square root of an integer
abs() - absolute value of an integer
2M
Any
four
correct
math
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22226
sin() - compute the sine value of an input value
cos()- compute the cosine value of an input value
pow()- compute the power of a input value
floor()- round down the input value
ceil()- round up the input value
function
with its
use ½M
each
(f)
Ans.
State the use of following symbols used for flowchart drawing:
(i) General processing
(ii) Decision making
(iii) Input/ Output statements
(iv) Start / Stop
2M
Correct
use of
symbols
½M
each
(g)
Ans.
State use of while loop with syntax.
While loop is used in programming to repeat a specific block of
statement until some end condition is met.
The syntax of a while loop is:
while (test Expression)
{
Statements…
statements….
}
2M
Use of
while
loop 1M
Syntax
of while
loop 1M
2.
(a)
Ans.
Attempt any THREE of the following:
Develop a simple ‘C’ program for addition and multiplication of
two integer numbers.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b,add,mul;
12
4M
Correct
Logic
2M
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22226
clrscr();
printf("Enter value for a and b:");
scanf("%d%d",&a,&b);
add=a+b;
mul=a*b;
printf("\nAddition of a and b=%d\n",add);
printf("\Multiplication of a and b=%d",mul);
getch();
}
Correct
syntax
2M
(b)
Ans.
Explain how to pass pointer to function with example.
(Note: Any other example showing pointer as a parameter in
function shall be considered)
When pointer (addresses) is passed to the function as an argument
instead of value then function is called as call by reference.
Example:
#include<stdio.h>
#include<conio.h>
int add(int *);
void main()
{
int *ptr,pos=0;
clrscr();
printf("Enter position:");
scanf("%d",&pos);
ptr=&pos;
printf("\nSum=%d",add(ptr));
getch();
}
int add(int *p)
{
int i=0;
int sum=0;
for(i=1;i<=(*p);i++)
{
sum=sum+i;
}
return sum;
}
4M
Explana
tion 2M
Example
2M
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In the above program function passes the address of „pos‟ to the ptr.
The value of ptr is passed while calling the function. In function
definition in *p it takes value of ptr instead of address for performing
addition of numbers up to specific position.
(c)
Ans.
Explain following functions:
getchar( )
putchar( )
getch( )
putch( )
with suitable examples.
getchar( ) -
It is function from stdio.h header file. This function is used to input a
single character.
The enter key is pressed which is followed by the character that is
typed. The character that is entered is echoed.
Syntax:
ch=getchar();
Example:
void main()
{
char ch;
ch = getchar();
printf("Input Char Is :%c",ch);
}
During the program execution, a single character gets or read
through the getchar(). The given value is displayed on the screen
and the compiler waits for another character to be typed. If you
press the enter key/any other characters and then only the given
character is printed through the printf function.
putchar( ) -
It is used from standard input (stdio.h) header file.
This function is the other side of getchar. A single character is
displayed on the screen.
Syntax:
putchar(ch);
void main()
{
4M
Explana
tion of
each
function
1M
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char ch='a';
putchar(ch);
getch();
}
getch( ) -
It is used from the console (conio.h) header file.
This function is used to input a single character.
The character is read instantly and it does not require an enter key to
be pressed.
The character type is returned but it does not echo on the screen.
Syntax:
ch=getch();
Where, ch - assigned the character that is returned by getch().
void main()
{
char ch;
ch = getch();
printf("Input Char Is :%c",ch);
}
During the program execution, a single character gets or read through
the getch(). The given value is not displayed on the screen and the
compiler does not wait for another character to be typed. And then,
the given character is printed through the printf function.
putch( )-
It is used from console input output header file (conio.h) This
function is a counterpart of getch().
Which means that it will display a single character on the screen.
The character that is displayed is returned.
Syntax:
putch(ch); Where, ch - the character that is to be printed.
void main()
{
char ch='a';
putch(ch)
}
(d)
Develop a program to accept an integer number and print
whether it is palindrome or not.
4M
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22226
Ans.
(Note: If string is considered instead of number for palindrome
checking, then that logic shall be considered)
#include<stdio.h>
#include<conio.h>
void main()
{
int n,num,rev=0,digit,i;
clrscr();
printf("Enter the number: ");
scanf("%d",&num);
n=num;
for(i=0;num!=0;++i)
{
digit=num%10;
rev=rev*10+digit;
num=num/10;
}
if(n==rev)
printf("Number is palindrome");
else
printf("Number is not palindrome");
getch();
}
Correct
Logic
2M
Correct
syntax
2M
3.
(a)
Ans.
Attempt any THREE of the following:
State the use of printf( ) & scanf( ) with suitable example.
printf( ) & scanf( ):
printf() and scanf() functions are library functions in C programming
language defined in “stdio.h”.
printf() function is used to print the character, string, float, integer,
octal and hexadecimal values onto the output screen.
scanf() function is used to read character, string, numeric data from
keyboard.
%d format specifier is used in printf() and scanf() to specify the value
of an integer variable.
%c is used to specify character,
%f for float variable, %s for string variable, and %x for hexadecimal
variable.
12
4M
Explana
tion of
printf,
scanf
1M each
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Example:
#include<stdio.h>
#include<conio.h>
void main() {
int i;
clrscr();
printf("Enter a number");
scanf("%d",&i);
printf("Entered number is: %d",i);
getch();
}
Example
2M
(b)
Ans.
Explain any four library functions under conio.h header file.
clrscr() -This function is used to clear the output screen.
getch() -It reads character from keyboard
getche()-It reads character from keyboard and echoes to o/p screen
putch - Writes a character directly to the console.
textcolor()-This function is used to change the text color
textbackground()-This function is used to change text background
4M
Any 4
function
1M each
(c)
Ans.
Explain how formatted input can be obtain, give suitable
example.
Formatted input:
When the input data is arranged in a specific format, it is called
formatted input. scanf function is used for this purpose in C.
General syntax:
scanf(“control string”, arg1, arg2..);
Control string here refers to the format of the input data. It includes
the conversion character %, a data type character and an optional
number that specifies the field width. It also may contain new line
character or tab. arg1, arg2 refers to the address of memory locations
where the data should be stored.
Example:
scanf(“%d”,&num1);
Eg:
#include<stdio.h>
#include<conio.h>
void main()
{
4M
Explana
tion 2M
Example
2M
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int i;
clrscr();
printf("Enter a number");
scanf("%d",&i);
printf("Entered number is: %d",i);
getch();
}
(d)
Ans.
Develop a program to find factorial of a number using recursion.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
int factorial(int num)
{
if(num==1)
{
return 1;
}
else
{
return(num*factorial(num-1));
}
}
void main() {
int num;
int result;
clrscr();
printf("Enter a number");
scanf("%d",&num);
result=factorial(num);
printf("Factorial of %d is %d",num,result);
getch();
}
4M
Correct
syntax
2M
Correct
logic 2M
4.
(a)
Ans.
Attempt any THREE of the following:
Write a program to sweep the values of variables a = 10, b = 5
using function.
(Note : Read sweep as swap in the question)
(Note: Any other logic using function shall be considered)
#include<stdio.h>
#include<conio.h>
12
4M
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void swapvalues(int *i, int *j)
{
int temp;
temp=*i;
*i=*j;
*j=temp;
}
void main() {
int a=10;
int b=5;
clrscr();
printf("The values before swaping:\na=%d, b=%d",a,b);
swapvalues(&a,&b);
printf("\nThe values after swaping:\na=%d, b=%d",a,b);
getch();
}
Correct
syntax
2M
Correct
logic 2M
(b)
Ans.
Develop a program using structure to print data of three students
having data members name, class, percentage.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
void main() {
struct student
{
char name[20];
char c[20];
int per;
} s[3];
int i;
clrscr();
for(i=0;i<3;i++)
{
printf("Enter name, class, percentage");
scanf("%s%s%d",&s[i].name,&s[i].c,&s[i].per);
}
for(i=0;i<3;i++)
{
printf("%s %s %d\n",s[i].name,s[i].c,s[i].per);
}
4M
Correct
syntax
2M
Correct
logic 2M
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22226
getch();
}
(c)
Ans.
Design a program to print a message 10 times.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
void main()
{
int i;
clrscr();
for(i=0;i<10;i++)
{
printf("Welcome to C programming\n");
}
getch();
}
4M
Correct
syntax
2M
Correct
logic 2M
(d)
Ans.
Draw a flowchart for checking whether given number is prime or
not.
(Note: Any correct flowchart shall be considered)
4M
Correct
symbols
1M
Correctn
ess of
flowchar
t 3M
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22226
(e)
Ans.
Implement a program to demonstrate logical AND operator.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
void main()
{
int i;
int j;
clrscr();
printf("Enter the values of i and j");
scanf("%d%d",&i,&j);
if(i==5 && j==5) {
printf("Both i and j are equal to 5");
} else {
printf("Both the values are different and either or both are not
equal to 5");
}
getch();
}
4M
Correct
Syntax
2M
Correct
logic 2M
5.
(a)
Ans.
Attempt any TWO of the following:
Draw a flowchart of Do-while loop and write a program to add
numbers until user enters zero.
Flowchart of Do-while loop:
12
6M
Correct
Flowcha
rt 3M
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22226
Program:-
#include<stdio.h>
#include<conio.h>
void main()
{
int no,sum=0;
clrscr();
do
{
printf("\n Enter a number:");
scanf("%d",&no);
sum=sum+no;
}while(no!=0);
printf("\n Sum of entered numbers =%d",sum);
getch();
}
Correct
program
3M
(b)
Ans.
Give a method to create, declare and initialize structure also
develop a program to demonstrate nested structure.
Declaration of structure:-
struct structure_name
{
data_type member 1;
data_type member 2;
.
.
.
data_type member n;
} structure variable 1, structure variable 2,..., structure variable n;
Example:-
struct student
{
int rollno;
char name[10];
}s1;
Initialization:-
struct student s={1,"abc"};
structure variable contains two members as rollno and name. the
6M
Creation
,
declarati
on 2M
Initializ
ation
1M
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above example initializes rollno to 1 and name to "abc".
Program:-
#include<stdio.h>
#include<conio.h>
struct college
{
int collegeid;
char collegename[20];
};
struct student
{
int rollno;
char studentname[10];
struct college c;
};
void main()
{
struct student s={1,"ABC",123,"Polytechnic"};
clrscr();
printf("\n Roll number=%d",s.rollno);
printf("\n Student Name=%s",s.studentname);
printf("\n College id=%d",s.c.collegeid);
printf("\n College name=%s",s.c.collegename);
getch();
}
Program
3M
(c)
Ans.
Implement a program to demonstrate concept of pointers to
function.
(Note: Any other relevant program shall be considered)
Pointer to function:
include<stdio.h>
int sum(int x, int y)
{
return x+y;
}
int main()
{
int s;
6M
Correct
logic 3M
Correct
syntax
3M
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int(*fp)(int, int);
fp = sum;
s = fp(10,12);
printf(“Sum = %d”,s);
return 0;
}
6.
(a)
Ans.
Attempt any TWO of the following:
Develop a program to swap two numbers using pointer and add
swaped numbers also print their addition.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
void swap(int *a,int *b)
{
int temp;
temp=*a;
*a=*b;
*b=temp;
}
void main()
{
int x,y,sum;
printf("\n Enter value for x:");
scanf("%d",&x);
printf("\n Enter value for y:");
scanf("%d",&y);
swap(&x,&y);
printf("\nx=%d",x);
printf("\ny=%d",y);
sum=x+y;
printf("Sum of x+y = %d",sum);
}
12
6M
Correct
logic for
swappin
g using
pointer
4M
Correct
logic for
addition
&
display
2M
(b)
Ans.
Design a programme in C to read the n numbers of values in an
array and display it in reverse order.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
#define max 50
void main()
{
6M
Correct
logic for
input
array
3M
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int a[max],i,n;
clrscr();
printf("\n Enter number of elements:");
scanf("%d",&n);
printf("\n Enter array element:");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("\n Array elements in reverse order:");
for(i=n-1;i>=0;i--)
printf("\t%d",a[i]);
getch();
}
Correct
logic to
display
in
reverse
3M
(c)
Ans.
Develop a program to find diameter, circumference and area of
circle using function.
(Note: Any other relevant logic shall be considered)
#include<stdio.h>
#include<conio.h>
void circle(float r)
{
float diameter,circumference,area;
diameter=2*r;
printf("\n Diameter=%f",diameter);
circumference=2*3.14*r;
printf("\n Circumference=%f",circumference);
area=3.14*r*r;
printf("\n Area=%f",area);
}
void main()
{
float radius;
clrscr();
printf("\n Enter radius:");
scanf("%f",&radius);
circle(radius);
getch();
}
6M
Correct
logic
using
function
to find
diameter
2M,circ
umferen
ce 2M,
area 2M
22226
[1 of 4] P.T.O.
11819
3 Hours / 70 Marks Seat No.
Instructions : (1) All Questions are compulsory.
(2) Illustrate your answers with neat sketches wherever necessary.
(3) Figures to the right indicate full marks.
(4) Assume suitable data, if necessary.
(5) Mobile Phone, Pager and any other Electronic Communication devices
are not permissible in Examination Hall.
(6) Preferably, write the answers in sequential order.
Marks
1. Attempt any FIVE of the following : 10
(a) Define Algorithm.
(b) Give the significance of <math.h> and <stdio.h> header files.
(c) Give syntax of if-else ladder.
(d) Define Array.
(e) Write syntax and use of pow( ) function of <math.h> header file.
(f) Define pointer. Write syntax for pointer declaration.
(g) Draw and label symbols used in flow chart.
22226 [2 of 4]
2. Attempt any THREE of the following : 12
(a) Write an algorithm to determine whether a given number is divisible by 5 or
not.
(b) Explain do – while loop with example.
(c) Explain one dimension and two dimension arrays.
(d) Write the output of following c program
#include<stdio.h>
int main ( )
{
char *ptr;
char str[]=“MAHARASHTRA STATE BOARD OF TECHNICAL
EDUCATION”;
ptr=str;
ptr=ptr+11;
printf(“%s”,++ptr);
return 0;
}
3. Attempt any THREE of the following : 12
(a) Explain increment and decrement operator.
(b) Explain User defined function with example.
(c) Explain conditional operator with example.
(d) Explain strlen( ) and strcpy( ) function with example.
22226 [3 of 4]
P.T.O.
4. Attempt any THREE of the following : 12
(a) Write algorithm and draw flow-chart to print even numbers from 1 to 100.
(b) Write a program to accept marks of four subjects as input from user. Calculate
and display total and percentage marks of student.
(c) Write a program to accept the value of year as input from the keyboard &
print whether it is a leap year or not.
(d) Write a program to accept a string as input from user and determine its length.
[Don’t use built in library function strlen( )]
(e) Write a program to swap two numbers using call by value.
5. Attempt any TWO of the following : 12
(a) Write a program using switch statement to check whether entered character is
VOWEL or CONSONANT.
(b) Write a program for addition of two 3 3 matrices.
(c) Write a program to Print values of variables and their addresses.
6. Attempt any TWO of the following : 12
(a) Write a program to declare structure employee having data member name,
age, street and city. Accept data for two employees and display it.
(b) If the value of a number (N) is entered through keyboard. Write a program
using recursion to calculate and display factorial of number (N).
(c) Write a program to accept two numbers from user and perform addition,
subtraction, multiplication and division operations using pointer.
_______________
22226 [4 of 4]
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Page 1 / 16
22226
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance
(Not applicable for subject English and Communication Skills).
4) While assessing figures, examiner may give credit for principal components indicated in the
figure. The figures drawn by candidate and model answer may vary. The examiner may give
credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model
answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant
answer based on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
Q.
No
.
Sub
Q.N.
Answer
Marking
Scheme
1.
(a)
Ans
Attempt any FIVE of the following:
Define Algorithm
Algorithm:- Algorithm is a stepwise set of instructions written to
perform a specific task.
10
2M
Correct
Definitio
n 2M
(b)
Ans
Give the significance of <math.h> and <stdio.h> header files.
“math.h” header file supports all the mathematical related functions
in C language.
stdio.h header file is used for input/output functions like scanf and
printf.
2M
Signific
ance of
each 1M
(c)
Ans
Give syntax of if-else ladder.
if(condition_expression_One)
{
statement1;
}
else if (condition_expression_Two)
{
statement2;
2M
Correct
syntax
2M
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22226
}
else if (condition_expression_Three)
{
statement3;
}
else
{
statement4;
}
(d)
Ans
Define Array.
An array is a collection of data items, all of the same type, accessed
using a common name.
A one-dimensional array consists of similar type of multiple values
in it.
A two dimensional array consists of row and column.
2M
Definitio
n of
array
2M
(e)
Ans
Write syntax and use of pow ()function of <math.h> header file.
pow()- compute the power of a input value
Syntax:
double pow (double x, double y);
2M
Syntax
and use
of pow()
1M each
(f)
Ans
Define pointer. Write syntax for pointer declaration.
Definition:
A pointer is a variable that stores memory address of another variable
which is of similar data type.
Declaration:
datatype *pointer_variable_name;
2M
Definitio
n of
pointer
1M,
Syntax
1M
(g)
Draw and label symbols used in flow chart.
2M
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Page 3 / 16
22226
Ans
Any
Four
Symbols
½ M
each
2.
(a)
Ans
Attempt any THREE of the following:
Write an algorithm to determine whether a given number is
divisible by 5 or not
Step 1- Start
Step 2- Read / input the number.
Step 3- if n%5==0 then goto step 5.
Step 4- else number is not divisible by 5 goto step 6.
Step 5- display the output number is divisible by 5.
Step 6- Stop
12
4M
Correct
algorith
m 4M
(b)
Ans
Explain do-while loop with example.
Do-While statement:
In some applications it is necessary to execute the body of the loop
before the condition is checked; such situation can be handled by
do statement.
At least once the body of loop will be executed.
do statement, first executes the body of the loop.
At the end of the loop, the test condition in the while statement is
evaluated. If the condition is true, then it continues to execute body
4M
Explana
tion 2M,
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22226
of the loop once again.
This process continues as long as the condition is true.
When the condition becomes false, the loops will be terminated
and the control goes to next statement after while statement.
Example:
#include <stdio.h>
#include <conio.h>
void main()
{
int i=1;
clrscr();
printf("\n Odd numbers from 1 to 20 are \n");
do
{
if(i%2 ! = 0)
printf("\n %d", i);
i++;
}while(i<=20); /* The loop iterates till the value of i is less than or
equal to 20 */
getch();
}
Any
relevant
Example
2M
(c)
Ans
Explain one dimension and two dimension arrays
i) One dimensional array:
An array is a collection of variables of the same type that are referred
through a common name. A specific element in an array is accessed
by an index. In C, all arrays consist of contiguous memory locations.
The lowest address corresponds to the first element and the highest
address to the last element.
Syntax: data_type array_name[array_size];
Example:
int marks[10];
ii) Two dimensional array :
Two dimensional array is a collection of similar type of data elements
arranged in the form of rows & columns.
Example:
Array can be declared as int arr[3][3]; In this there can be 9 elements
in an array with 3 rows and 3 columns.
4M
Explana
tion of
one
dimensi
onal and
two
dimensi
onal
array
2M each
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22226
(d)
Ans
Write the output of following c program
#include<stdio.h>
int main()
{
char *ptr;
char str[]=”MAHARASHTRA STATE BOARD OF
TECHNICAL EDUCATION”;
ptr=str;
ptr=ptr+11;
printf(“%s”, ++ptr);
return 0;
}
Output :
STATE BOARD OF TECHNICAL EDUCATION
4M
Correct
output
4M
3
(a)
Ans
Attempt any THREE of the following:
Explain increment and decrement operator.
Increment operator is used to increment or increase the value of a
variable by one. It is equivalent to adding one to the value of the
variable. The symbol used is ++. The decrement operator is used to
decrement or decrease the value of variable by 1. It is equivalent to
subtracting one from the value of the variable. The symbol used is --.
Syntax: ++var or var++ for increment and --var or var--for
decrement.
Example:
int m=5;
int n = ++m;
printf(%d%d”,m,n);
When the increment operator is used prior to the variable name m, the
value of the variable m is incremented first and then assigned to the
variable n. The values of both the variable m and n here will be 6. But
if the increment operator ++ is used after the variable name, then the
value of the variable m is assigned to the variable n and then the
value of m is increased. Therefore the values of m and n will be 6 and
5 respectively.
Example for decrement operator
int m=5;
int n=--m;
12
4M
Explana
tion of
each
2M
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22226
printf(“%d%d”,m,n);
or
#include<stdio.h>
#include<conio.h>
void main() {
int m=4,n=6;
clrscr();
printf("values of m and n before changing%d%d",m,n);
m++;
n--;
printf("\nvalues after changing%d%d",m,n);
getch();
}
(b)
Ans
Explain User defined function with example.
Functions are basic building blocks in a program. It can be
predefined/ library functions or user defined functions. Predefined
functions are those which are already available in C library. User
defined functions are those which are written by the users to complete
a specific task. Execution of a C program starts from main(). User
defined functions should be called from main() for it to execute. A
user defined function has a return type and a name. it my or may not
contain parameters.
The general syntax of a user defined function :
Return_type func_name(parameter list)
Example:
#include<stdio.h>
#include<conio.h>
void myFunc(int a) {
printf("The value is: %d",a);
}
void main() {
myFunc(10);
getch()
}
4M
Explana
tion with
general
syntax
2M
Example
2M
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22226
(c)
Ans
Explain conditional operator with example.
Conditional operators return one value if condition is true and returns
another value is condition is false. This operator is also called as
ternary operator as it takes three arguments.
Syntax :
(Condition? true_value: false_value);
Example:
#include<stdio.h>
#include<conio.h>
void main() {
int i;
clrscr();
printf("Enter a number:");
scanf("%d",&i);
i%2==0?printf("%d is even",i):printf("%d is odd",i) ;
getch();
}
4M
Explana
tion 2M
Example
2M
(d)
Ans
Explain strlen() and strcpy() function with example.
strlen()- this function is used to find the length of a string. It counts
the number of characters comprising the string.
Syntax:
strlen(char[] str)- finds the length of the string str.
Example:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main() {
char str[] = "mystring";
int len=0;
clrscr();
len=strlen(str);
printf("Length of string is :%d",len);
getch();
}
Strcpy()– this function is used to copy the contents of a string to
other.
4M
Explan
ation &
Exampl
e of
each
2M
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Syntax:
strcpy(char[] dest, char[] source)- copies the contents of the string
source to destination.
Example:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main() {
char source[]="mystring";
char dest[10];
clrscr();
printf("%s%s",source,dest);
strcpy(dest,source);
printf("\n%s %s",source, dest);
getch();
}
4
(a)
Ans
Attempt any THREE of the following
Write algorithm and draw flow-chart to print even numbers
from 1 to 100.
Algorithm
1. Start
2. Initialize the variable i to 1.
3. while i<=100
4. if i%2==0
5. print the number
6. increment value of i
7. stop
12
4M
Algorith
m 2M
Flowcha
rt 2M
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22226
Flowchart
(b)
Ans
Write a program to accept marks of four subjects as input from
user. Calculate and display total and percentage marks of
student.
Note: Any other correct logic shall be considered
#include<stdio.h>
#include<conio.h>
void main() {
float marks[4];
float total=0.0, perc=0.0;
int i;
4M
Relevant
logic
2M
Syntax
2M
start
Initialize variable i=1
Is i <=100?
Is i%2==0?
Print i
i=i+1
NO
NO
YES
stop
YES
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22226
clrscr();
for(i=1;i<=4;i++) {
printf("Enter marks of subject %d",i);
scanf("%f%",&marks[i]);
}
for(i=1;i<=4;i++){
total=total+marks[i];
}
printf("Total is :%f",total);
perc=total/4;
printf("Percentage is %f",perc);
getch();
}
(c)
Ans
Write a program to accept the value of year as input from the
keyboard & print whether it is a leap year or not.
#include<stdio.h>
#include<conio.h>
void main() {
int year;
clrscr();
printf("Enter year");
scanf("%d",&year);
if(year%4==0) {
printf("Year %d is a leap year",year);
} else {
printf("Year %d is not a leap year",year);
}
getch();
}
4M
Correct
Logic
2M
Correct
Syntax
2M
(d)
Ans
Write a program to accept a string as input from user and
determine its length. [Don’t use built in library function strlen()]
#include<stdio.h>
#include<conio.h>
void main(){
char str[50];
int i, len=0;
clrscr();
printf("Enter a string");
4M
Correct
Logic
2M
Correct
Syntax
2M
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22226
scanf("%s",&str);
for(i=0; str[i]!='\0'; i++){
len++;
}
printf("The length of string is %d",len);
getch();
}
(e)
Ans
Write a program to swap two numbers using call be value.
#include<stdio.h>
#include<conio.h>
void swap(int a, int b) {
int temp;
temp=a;
a=b;
b=temp;
printf("Numbers after swapping no1=%d and no2=%d",a,b);
}
void main() {
int no1, no2;
clrscr();
printf("Enter the 2 numbers");
scanf("%d%d",&no1,&no2);
printf("Numbers before swapping no1=%d and no2= %d",no1, no2);
swap(no1,no2);
getch();
}
4M
Correct
Logic
2M
Correct
Syntax
2M
5
(a)
Ans
Attempt any TWO of the following:
Write a program using switch statement to check whether
entered character is VOWEL or CONSONANT
Note : Assume that the entered character is only alphabet.
#include<stdio.h>
#include<conio.h>
void main()
{
char ch;
clrscr();
printf("Enter character:");
scanf("%c",&ch);
12
6M
characte
r input-
2M
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switch(ch)
{
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
printf("\n Entered character is VOWEL");
break;
default:
printf("\n Entered character is CONSONANT");
}
getch();
}
Display
vowel-
2M
Display
consona
nt
2M
(b)
Ans
Write a program for addition of two 3 x 3 matrices.
#include<stdio.h>
#include<conio.h>
void main()
{
int a[3][3],b[3][3],c[3][3],i,j;
clrscr();
printf("Enter first matrix elements:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("\nEnter second matrix elements:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
6M
Input of
two
matrices
2M
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{
scanf("%d",&b[i][j]);
}
}
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
c[i][j]=a[i][j]+b[i][j];
}
}
printf("\n\nAddition of two matrices is:");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
printf("%d\t",c[i][j]);
}
}
getch();
}
Addition
of
matrices
2M
Display
of
addition
2M
(c)
Ans
Write a program to Print values of variables and their addresses.
Note : 1) Variables can be of any data type.
2)Use of & or pointer to display address shall be
considered.
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b;
clrscr();
a=5;
b=10;
printf("\n Value of a=%d",a);
printf("\n Address of a=%u",&a);
printf("\n Value of b=%d",b);
printf("\n Address of b=%u",&b);
getch();
}
6M
Display
values
of
variable-
3M
Display
address
of
variable
3M
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6
(a)
Ans
Attempt any TWO of the following:
Write a program to declare structure employee having data
member name, age, street and city. Accept data for two
employees and display it.
Note : Two structure variables or array of structure variables shall
be considered.
#include<stdio.h>
#include<conio.h>
struct employee
{
char name[10],street[10],city[10];
int age;
};
void main()
{
int i;
struct employee e[2];
clrscr();
for(i=0;i<2;i++)
{
printf("\n Enter name:");
scanf("%s",&e[i].name);
printf("\n Enter age:");
scanf("%d",&e[i].age);
printf("\n Enter street:");
scanf("%s",&e[i].street);
printf("\n Enter city:");
scanf("%s",&e[i].city);
}
for(i=0;i<2;i++)
{
printf("\n Name=%s",e[i].name);
printf("\n Age=%d",e[i].age);
printf("\n Street=%s",e[i].street);
printf("\n City=%s",e[i].city);
}
getch();
}
12
6M
Declarat
ion of
structur
e-2M
Acceptin
g data-
2M
Displayi
ng
data2M
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(b)
Ans
If the value of a number (N) is entered through keyboard. Write
a program using recursion to calculate and display factorial of
number (N).
#include<stdio.h>
#include<conio.h>
int factorial(int N);
void main()
{
int N,fact;
clrscr();
printf("Enter number:");
scanf("%d",&N);
fact=factorial(N);
printf("\n Factorial is:%d",fact);
getch();
}
int factorial(int N)
{
if(N==1)
return(1);
else
return(N*factorial(N-1));
}
6M
Main
function
definitio
n-3M,
Recursiv
e
function
definitio
n-3M
(c)
Ans
Write a program to accept two numbers from user and perform
addition, subtraction, multiplication and division operations
using pointer.
#include<stdio.h>
#include<conio.h>
void main()
{
int no1,no2,*ptr1,*ptr2,result;
clrscr();
printf("Enter no1:");
scanf("%d",&no1);
printf("\nEnter no2:");
scanf("%d",&no2);
ptr1=&no1;
ptr2=&no2;
result=*ptr1+*ptr2;
6M
Acceptin
g
numbers
1M
Pointer
initializa
tion-1M
Addition
1M
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printf("\n Addition=%d",result);
result=*ptr1-*ptr2;
printf("\n Subtraction=%d",result);
result=*ptr1**ptr2;
printf("\n Multiplication=%d",result);
result=*ptr1/(*ptr2);
printf("\n Division=%d",result);
getch();
}
subtracti
on-1M
multiplic
ation-
1M
division-
1M
22226
[1 of 2] P.T.O.
21819
3 Hours / 70 Marks Seat No.
Instructions : (1) All Questions are compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Use of Non-programmable Electronic Pocket Calculator is permissible.
Marks
1. Attempt any FIVE of the following : 10
(a) Draw flowchart for checking whether given number is even or odd.
(b) List any four keywords used in ‘C’ with their use.
(c) Write the syntax of switch case statement.
(d) State any two differences between while and do-while statement.
(e) State difference between array and string.
(f) Declare a structure student with element roll-no and name.
(g) Distinguish between call by value and call by reference.
2. Attempt any THREE of the following : 12
(a) State four arithmetic operations perform on pointer with example.
(b) Draw flowchart for checking weather given number is prime or not.
(c) Write a program to reverse the number 1234 (i.e. 4321) using function.
(d) Differentiate between character array and integer array with respect to size
and initialisation.
22226 [2 of 2]
3. Attempt any THREE of the following : 12
(a) Write a program to sum all the odd numbers between 1 to 20.
(b) Explain any four bit-wise operator used in ‘C’ with example.
(c) With suitable example, explain how two dimensional arrays can be created.
(d) Explain any two string functions with example.
4. Attempt any THREE of the following : 12
(a) Draw flowchart for finding largest number among three numbers.
(b) Describe generic structure of ‘C’ program.
(c) Write a program to take input as a number and reverse it by while loop.
(d) Write a program to accept 10 numbers in array and arrange them in ascending
order.
(e) Explain meaning of following statement with reference to pointers :
int *a, b;
b = 20;
*a = b;
A = &b;
5. Attempt any TWO of the following : 12
(a) Write a program to perform addition, subtraction, multiplication and division
of two integer number using function.
(b) Define Array. Write a program to accept ten numbers in array. Sort array
element and display.
(c) Write a program to print reverse of a entered string using pointer.
6. Attempt any TWO of the following : 12
(a) Explain recursion with suitable example. List any two advantages.
(b) Write a program to accept ten numbers and print average of it.
(c) Enlist different format specifiers with its use.
_______________
MA
H
ARA
SH
T
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
I
m
p
orta
n
t
Ins
tr
u
ct
i
o
ns
to exam
in
er
s
:
1) The an
s
wer
s
s
hou
l
d be exa
mi
ned by key word
s
and no
t
a
s
word-
t
o-word a
s
g
i
ven
i
n
t
he
m
ode
l
an
s
wer
s
che
m
e.
2) The
m
ode
l
an
s
wer and
t
he an
s
wer wr
itt
en by cand
i
da
t
e
m
ay vary bu
t
t
he exa
mi
ner
m
ay
t
ry
t
o
a
ss
e
ss
t
he under
st
and
i
ng
l
eve
l
of
t
he cand
i
da
t
e.
3) The
l
anguage error
s
s
uch a
s
gra
mm
a
ti
ca
l
,
s
pe
lli
ng error
s
s
hou
l
d no
t
be g
i
ven
m
ore I
m
por
t
ance
(No
t
app
li
cab
l
e for
s
ub
j
ec
t
Eng
lis
h and
C
o
mm
un
i
ca
ti
on
S
k
ills
).
4) Wh
il
e a
ss
e
ssi
ng f
i
gure
s
, exa
mi
ner
m
ay g
i
ve cred
it
for pr
i
nc
i
pa
l
co
m
ponen
ts
i
nd
i
ca
t
ed
i
n
t
he
f
i
gure. The f
i
gure
s
drawn by cand
i
da
t
e and
m
ode
l
an
s
wer
m
ay vary. The exa
mi
ner
m
ay g
i
ve
cred
it
for any equ
i
va
l
en
t
f
i
gure drawn.
5)
C
red
its
m
ay be g
i
ven
st
ep w
is
e for nu
m
er
i
ca
l
prob
l
e
ms
. In
s
o
m
e ca
s
e
s
,
t
he a
ss
u
m
ed con
st
an
t
an
s
wer.
6) In ca
s
e of
s
o
m
e que
sti
on
s
cred
it
m
ay be g
i
ven by
j
udge
m
en
t
on par
t
of exa
mi
ner of re
l
evan
t
d
i
ng.
7)
F
or progra
mmi
ng
l
anguage paper
s
, cred
it
m
ay be g
i
ven
t
o any o
t
her progra
m
ba
s
ed on
equ
i
va
l
en
t
concep
t
.
Q.
No
.
S
ub
Q.N.
An
s
wer
M
ark
i
ng
S
che
m
e
1.
(a)
A
ns
.
Attem
p
t a
n
y F
I
V
E
of t
h
e fo
ll
ow
in
g:
Draw f
l
owc
h
art for c
h
ec
kin
g w
h
et
h
er g
i
ve
n
nu
m
b
er
is
eve
n
or
o
dd
.
10
2M
C
o
rr
ec
t
l
og
i
c 1
M
R
e
l
eva
nt
s
y
m
bo
l
1
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
(
b
)
A
ns
.
( ).
Keyword U
s
e
au
t
o I
t
is
u
s
ed
t
o dec
l
are au
t
o
st
orage c
l
a
ss
var
i
ab
l
e.
break I
t
is
u
s
ed
t
o ex
it
fro
m
b
l
ock or
l
oop.
ca
s
e I
t
is
u
s
ed
t
o repre
s
en
t
po
ssi
b
l
e ca
s
e
i
n
si
de
s
w
it
ch
ca
s
e
st
a
t
e
m
en
t
char U
s
ed for dec
l
ara
ti
on of charac
t
er
t
ype var
i
ab
l
e
con
st
I
t
is
u
s
ed
t
o dec
l
are a con
st
an
t
.
con
ti
nue I
t
is
u
s
ed pa
ss
con
t
ro
l
a
t
t
he beg
i
nn
i
ng of
t
he
l
oop
defau
lt
I
t
is
u
s
ed
t
o repre
s
en
t
defau
lt
ca
s
e
i
n
si
de
s
w
it
ch
ca
s
e
st
a
t
e
m
en
t
.
do I
t
is
u
s
ed
t
o execu
t
e
l
oop
i
n a
ss
oc
i
a
ti
on w
it
h
wh
il
e cond
iti
on.
doub
l
e U
s
ed for dec
l
ara
ti
on of doub
l
e
t
ype var
i
ab
l
e
e
ls
e I
t
is
u
s
ed w
it
h
i
f
st
a
t
e
m
en
t
t
o
t
ran
s
fer con
t
ro
l
t
o
st
a
t
e
m
en
t
when cond
iti
on
is
fa
ls
e.
enu
m
I
t
is
u
s
ed
t
o dec
l
are enu
m
era
t
ed da
t
a.
ex
t
ern I
t
is
u
s
ed
t
o dec
l
are ex
t
ern
st
orage c
l
a
ss
var
i
ab
l
e
f
l
oa
t
U
s
ed for dec
l
ara
ti
on of f
l
oa
t
t
ype var
i
ab
l
e
for U
s
ed for repe
titi
ve execu
ti
on of
st
a
t
e
m
en
ts
go
t
o I
t
is
u
s
ed
t
o
t
ran
s
fer con
t
ro
l
fro
m
one
st
a
t
e
m
en
t
t
o ano
t
her
i
f I
t
is
u
s
ed for cond
iti
on check
i
ng
i
n
t
U
s
ed for dec
l
ara
ti
on of
i
n
t
eger
t
ype var
i
ab
l
e
l
ong U
s
ed for dec
l
ara
ti
on of
l
ong
t
ype var
i
ab
l
e
reg
ist
er I
t
is
u
s
ed
t
o dec
l
are reg
ist
er
st
orage c
l
a
ss
var
i
ab
l
e
re
t
urn I
t
is
u
s
ed
t
o re
t
urn va
l
ue fro
m
func
ti
on.
s
hor
t
U
s
ed for dec
l
ara
ti
on of
s
hor
t
t
ype var
i
ab
l
e
si
gned U
s
ed for dec
l
ara
ti
on of
si
gned
t
ype var
i
ab
l
e
si
zeof I
t
re
t
urn
s
m
e
m
ory
si
ze a
ll
oca
t
ed
t
o var
i
ab
l
e or
da
t
a
t
ype
st
a
ti
c I
t
is
u
s
ed
t
o dec
l
are
st
a
ti
c
st
orage c
l
a
ss
var
i
ab
l
e
st
ruc
t
I
t
is
u
s
ed
t
o dec
l
are u
s
er def
i
ned da
t
a
t
ype
st
ruc
t
ure
2M
An
y
fo
ur
key
w
o
r
d
s
1
M
U
s
e 1
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
s
w
it
ch I
t
is
u
s
ed
t
o
m
ake dec
isi
on fro
m
m
u
lti
p
l
e nu
m
ber
of
i
npu
ts
t
ypedef U
s
ed
t
o redef
i
ne
t
he na
m
e of an ex
isti
ng var
i
ab
l
e
t
ype.
un
i
on I
t
is
u
s
ed
t
o dec
l
are
t
he da
t
a
t
ype un
i
on
un
si
gned U
s
ed for dec
l
ara
ti
on of un
si
gned
t
ype var
i
ab
l
e
vo
i
d
S
pec
i
fy
t
ha
t
func
ti
on doe
s
no
t
re
t
urn any va
l
ue
vo
l
a
til
e I
t
is
u
s
ed
t
o dec
l
are a vo
l
a
til
e var
i
ab
l
e
wh
il
e U
s
ed for repe
titi
ve execu
ti
on of
st
a
t
e
m
en
ts
(c)
A
ns
.
Wr
i
te t
h
e
s
y
n
tax of
s
w
i
tc
h
ca
s
e
s
tateme
n
t.
s
w
it
ch(var
i
ab
l
e)
{
ca
s
e va
l
ue1
:
st
a
t
e
m
en
ts
break
;
ca
s
e va
l
ue2
:
st
a
t
e
m
en
ts;
break
;
.
.
.
defau
lt:
st
a
t
e
m
en
ts;
break
;
}
2M
C
o
rr
ec
t
s
y
nt
ax
2
M
(
d
)
A
ns
.
S
tate a
n
y two
di
ffere
n
ce
s
b
etwee
n
w
hil
e a
nd
d
o-w
hil
e
s
tateme
n
t.
(No
t
e:
An
y 2 po
ints
sh
a
ll
be co
nsi
de
r
ed).
w
hil
e Do-w
hil
e
In 'wh
il
e'
l
oop
t
he con
t
ro
lli
ng
cond
iti
on appear
s
a
t
t
he
st
ar
t
of
t
he
l
oop.
In 'do-wh
il
e'
l
oop
t
he
con
t
ro
lli
ng cond
iti
on appear
s
a
t
t
he end of
t
he
l
oop.
The
it
era
ti
on
s
do no
t
occur
i
f,
t
he cond
iti
on a
t
t
he f
i
r
st
it
era
ti
on, appear
s
fa
ls
e.
The
it
era
ti
on occur
s
a
t
l
ea
st
once even
i
f
t
he cond
iti
on
is
fa
ls
e a
t
t
he f
i
r
st
it
era
ti
on.
I
t
is
an en
t
ry con
t
ro
ll
ed
l
oop I
t
is
an ex
it
con
t
ro
ll
ed
l
oop
wh
il
e(cond
iti
on) {
body
do {
body
2M
An
y
tw
o
d
i
ffe
r
e
n
ce
s
1
M
eac
h
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
} }wh
il
e(cond
iti
on)
;
(e)
A
ns
.
S
tate
di
ffere
n
ce
b
etwee
n
array a
nd
s
tr
in
g.
(No
t
e:
An
y
tw
o va
li
d po
ints
sh
a
ll
be co
nsi
de
r
ed).
Array
S
tr
in
g
Array can be of any
t
ype
li
ke
i
n
t
, f
l
oa
t
, char.
St
r
i
ng can con
t
a
i
n on
l
y
charac
t
er
s
.
E
l
e
m
en
t
E
l
e
m
en
ts
i
n an array
can be acce
ss
ed u
si
ng
its
po
siti
on
li
ke a[2].
s
i
n an array
can be acce
ss
ed u
si
ng
its
po
siti
on
li
ke a[2].
C
harac
t
er
s
i
n
st
r
i
ng are acce
ss
ed
s
equen
ti
a
ll
y fro
m
f
i
r
st
t
o
l
a
st
.
Array doe
s
no
t
end w
it
h a nu
ll
charac
t
er
S \
charac
t
er.
Array
si
ze once dec
l
ared canno
t
be changed
St
r
i
ng
si
ze can be
m
od
i
f
i
ed
u
si
ng po
i
n
t
er.
2M
An
y
tw
o
po
ints
1
M
fo
r
eac
h
(f)
A
ns
.
Dec
l
are a
s
tr
u
ct
u
re
s
t
ud
e
n
t w
i
t
h
e
l
eme
n
t ro
ll
-
n
o a
nd
n
ame.
st
ruc
t
st
uden
t
{
i
n
t
ro
ll
_no
;
char na
m
e[20]
;
}
;
2M
C
o
rr
ec
t
dec
l
a
r
a
ti
o
n
2
M
(g)
A
ns
.
D
is
t
in
g
uish
b
etwee
n
ca
ll
b
y va
lu
e a
nd
ca
ll
b
y refere
n
ce.
(No
t
e:
An
y
tw
o po
ints
sh
a
ll
be co
nsi
de
r
ed).
Ca
ll
b
y va
lu
e Ca
ll
b
y refere
n
ce
A copy of ac
t
ua
l
argu
m
en
ts
is
pa
ss
ed
t
o re
s
pec
ti
ve for
m
a
l
argu
m
en
ts
.
The addre
ss
of ac
t
ua
l
argu
m
en
ts
is
pa
ss
ed
t
o for
m
a
l
argu
m
en
ts
Ac
t
ua
l
argu
m
en
ts
w
ill
re
m
a
i
n
s
afe,
t
hey canno
t
be
m
od
i
f
i
ed
acc
i
den
t
a
ll
y.
A
lt
era
ti
on
t
o ac
t
ua
l
argu
m
en
ts
is
po
ssi
b
l
e w
it
h
i
n fro
m
ca
ll
ed
func
ti
on
;
t
herefore
t
he code
m
u
st
hand
l
e argu
m
en
ts
carefu
ll
y e
ls
e
you ge
t
unexpec
t
ed re
s
u
lts
.
Addre
ss
of
t
he ac
t
ua
l
and for
m
a
l
argu
m
en
ts
are d
i
fferen
t
Addre
ss
of
t
he ac
t
ua
l
and for
m
a
l
argu
m
en
ts
are
t
he
s
a
m
e
C
hange
s
m
ade
i
n
si
de
t
he func
ti
on
is
no
t
ref
l
ec
t
ed
i
n o
t
her func
ti
on
s
C
hange
s
m
ade
i
n
t
he func
ti
on
is
ref
l
ec
t
ed ou
tsi
de a
ls
o.
2M
An
y
tw
o
po
ints
1
M
eac
h
2.
(a)
Attem
p
t a
n
y
TH
R
EE
of t
h
e fo
ll
ow
in
g:
S
tate fo
u
r ar
i
t
h
met
i
c o
p
erat
i
o
ns
p
erform o
n
p
o
in
ter w
i
t
h
12
4M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
A
ns
.
exam
pl
e.
(No
t
e:
C
ode
sni
ppe
t
sh
a
ll
be co
nsi
de
r
ed)
The po
i
n
t
er ar
it
h
m
e
ti
c
is
done a
s
per
t
he da
t
a
t
ype of
t
he po
i
n
t
er. The
ba
si
c opera
ti
on
s
on po
i
n
t
er
s
are
In
creme
n
t:
I
t
is
u
s
ed
t
o
i
ncre
m
en
t
t
he po
i
n
t
er. Each
tim
e a po
i
n
t
er
is
i
ncre
m
en
t
ed,
it
po
i
n
ts
t
o
t
he nex
t
l
oca
ti
on w
it
h re
s
pec
t
t
o
m
e
m
ory
si
ze .
Exa
m
p
l
e,
If p
t
r
is
an
i
n
t
eger po
i
n
t
er
st
ored a
t
addre
ss
1000,
t
hen p
t
r
++
s
how
s
1002 a
s
i
ncre
m
en
t
ed
l
oca
ti
on for an
i
n
t
.I
t
i
ncre
m
en
ts
by
t
wo
l
oca
ti
on
s
a
s
it
requ
i
re
s
t
wo by
t
e
s
st
orage.
Decreme
n
t:
I
t
is
u
s
ed
t
o decre
m
en
t
t
he po
i
n
t
er. Each
tim
e a po
i
n
t
er
is
decre
m
en
t
ed,
it
po
i
n
ts
t
o
t
he prev
i
ou
s
l
oca
ti
on w
it
h re
s
pec
t
t
o
m
e
m
ory
si
ze.
Exa
m
p
l
e,
If
t
he curren
t
po
siti
on of po
i
n
t
er
is
1002,
t
hen decre
m
en
t
opera
ti
on
p
t
r-- re
s
u
lts
i
n
t
he po
i
n
t
er po
i
n
ti
ng
t
o
t
he
l
oca
ti
on 1000
i
n ca
s
e of
i
n
t
eger po
i
n
t
er a
s
it
requ
i
re
t
wo by
t
e
s
st
orage.
A
ddi
t
i
o
n
When add
iti
on opera
ti
on
is
perfor
m
ed on po
i
n
t
er,
it
g
i
ve
s
t
he
l
oca
ti
on
i
ncre
m
en
t
ed by
t
he added va
l
ue accord
i
ng
t
o da
t
a
t
ype.
Eg
:
If p
t
r
is
an
i
n
t
eger po
i
n
t
er
st
ored a
t
addre
ss
1000,
Then p
t
r
+
2
s
how
s
1000
+
(2*2)
=
1004 a
s
i
ncre
m
en
t
ed
l
oca
ti
on for an
i
n
t
.
Sub
tract
i
o
n
When
s
ub
t
rac
ti
on opera
ti
on
is
perfor
m
ed on
t
he po
i
n
t
er var
i
ab
l
e,
it
g
i
ve
s
t
he
l
oca
ti
on decre
m
en
t
ed by
t
he
s
ub
t
rac
t
ed va
l
ue accord
i
ng
t
o
da
t
a
t
ype.
Eg
:
If p
t
r
is
an
i
n
t
eger po
i
n
t
er
st
ored a
t
addre
ss
1004,
Then p
t
r-2
s
how
s
1004-(2*2)
=
1000 a
s
decre
m
en
t
ed
l
oca
ti
on for an
i
n
t
.
E
ac
h
ope
r
a
ti
o
n
with
exa
m
p
l
e
1
M
(
b
) Draw f
l
owc
h
art for c
h
ec
kin
g weat
h
er g
i
ve
n
nu
m
b
er
is
p
r
i
me or 4M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
A
ns
.
n
ot.
C
o
rr
ec
t
l
og
i
c 2
M
S
y
m
bo
ls
2
M
(c)
A
ns
.
Wr
i
te a
p
rogram to rever
s
e t
h
e
nu
m
b
er 1234 (
i
.e. 4321)
usin
g
f
un
ct
i
o
n
.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d f
i
nd
R
ever
s
e()
;
vo
i
d
m
a
i
n()
{
f
i
nd
R
ever
s
e()
;
}
vo
i
d f
i
nd
R
ever
s
e()
{
4M
C
o
rr
ec
t
s
y
nt
ax
2
M
C
o
rr
ec
t
l
og
i
c 2
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
i
n
t
nu
m
, re
s=
0,an
s=
0
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er
t
he nu
m
ber")
;
s
canf("%d",
&
nu
m
)
;
wh
il
e(nu
m
!
=
0)
{
re
s=
nu
m
%10
;
an
s=
an
s
*10
+
re
s;
nu
m=
nu
m/
10
;
}
pr
i
n
t
f("
R
ever
s
e nu
m
ber
is
%d", an
s
)
;
ge
t
ch()
;
}
(
d
)
A
ns
.
D
i
ffere
n
t
i
ate
b
etwee
n
c
h
aracter array a
nd
in
teger array w
i
t
h
re
sp
ect to
si
ze a
nd
ini
t
i
a
lis
at
i
o
n
.
Parameter C
h
aracter Array
In
teger Array
Si
ze La
st
l
oca
ti
on
i
n
charac
t
er array
is
f
ill
ed w
it
h '
\
0'
s
o
t
he
array
si
ze
s
hou
l
d be
s
o dec
l
ared
t
ha
t
it
s
hou
l
d have one
l
a
st
l
oca
ti
on for '
\
0'
charac
t
er.
No ex
t
ra
l
oca
ti
on
t
han
t
he nu
m
ber of
e
l
e
m
en
ts
is
requ
i
red.
In
iti
a
li
za
ti
on In
iti
a
li
za
ti
on can be
done
li
ke
:
char
st
r[4]
=
{'a','b','c','
\
0'}
;
char
st
r[4]
=
"abc"
;
In
iti
a
li
za
ti
on can be
done
li
ke
:
i
n
t
arr[4]
=
{1,2,3,4}
;
4M
E
ac
h
pa
r
a
m
e
t
e
r
2
M
3.
(a)
A
ns
.
Attem
p
t a
n
y
TH
R
EE
of t
h
e fo
ll
ow
in
g:
Wr
i
te a
p
rogram to
su
m a
ll
t
h
e o
dd
nu
m
b
er
s
b
etwee
n
1 to 20.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
i
n
t
s
u
m=
0,
i;
c
l
r
s
cr()
;
for(
i=
1
;i<=
20
;i++
)
12
4M
C
o
rr
ec
t
l
og
i
c
2
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
{
i
f(
i
%2
==
1)
s
u
m=s
u
m+i;
}
ge
t
ch()
;
}
C
o
rr
ec
t
s
y
nt
ax
2
M
(
b
)
A
ns
.
E
x
pl
a
in
a
n
y fo
u
r
bi
t-
Bi
tw
is
e o
p
erator
s
:
Bi
tw
is
e
O
R
|
I
t
t
ake
s
2 b
it
pa
tt
ern
s
and perfor
ms
O
R
opera
ti
on
s
on each pa
i
r of
corre
s
pond
i
ng b
its
. The fo
ll
ow
i
ng exa
m
p
l
e w
ill
exp
l
a
i
n
it
.
1010
1100
--------
O
R
1110
Bi
tw
is
e AND &
I
t
t
ake
s
2 b
it
pa
tt
ern
s
and perfor
ms
AND opera
ti
on
s
w
it
h
it
.
1010
1100
-------
AND 1000
-------
The
Bit
w
is
e AND w
ill
t
ake pa
i
r of b
its
fro
m
each po
siti
on, and
i
f
on
l
y bo
t
h
t
he b
it
is
1,
t
he re
s
u
lt
on
t
ha
t
po
siti
on w
ill
be 1.
Bit
w
is
e
AND
is
u
s
ed
t
o Turn-Off b
its
.
Bi
tw
is
e N
OT
One
s
co
m
p
l
e
m
en
t
opera
t
or (
Bit
w
is
e NOT)
is
u
s
ed
t
o conver
t
each
-b
it
t
o 0- -b
it
t
o1-
unary opera
t
or
i
.e.
it
t
ake
s
on
l
y one operand.
1001
NOT 0110
-------
Bi
tw
is
e X
O
R
^
Bit
w
is
e XO
R
^,
t
ake
s
2 b
it
pa
tt
ern
s
and perfor
m
XO
R
opera
ti
on w
it
h
it
.
4M
E
xp
l
a
n
a
ti
o
n
with
exa
m
p
l
e
of a
n
y
fo
ur
b
itwis
e
ope
r
a
t
o
r
1
M
eac
h
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
0101
0110
------
XO
R
0011
------
L
eft
shi
ft
Op
erator
<<
The
l
ef
t
s
h
i
f
t
opera
t
or w
ill
s
h
i
f
t
t
he b
its
t
oward
s
l
ef
t
for
t
he g
i
ven
nu
m
ber of
tim
e
s
.
i
n
t
a
=
2
<<
1
;
R
i
g
h
t
shi
ft
Op
erator
>>
The r
i
gh
t
s
h
i
f
t
opera
t
or w
ill
s
h
i
f
t
t
he b
its
t
oward
s
r
i
gh
t
for
t
he g
i
ven
nu
m
ber of
tim
e
s
.
i
n
t
a
=
8
>>
1
;
(c)
A
ns
.
W
i
t
h
sui
ta
bl
e exam
pl
e, ex
pl
a
in
h
ow two
di
me
nsi
o
n
a
l
array
s
ca
n
b
e create
d
.
The array wh
i
ch
is
u
s
ed
t
o repre
s
en
t
and
st
ore da
t
a
i
n a
t
abu
l
ar for
m
is
ca
ll
ed a
s
t
wo d
im
en
si
ona
l
array.
S
uch
t
ype of array
is
s
pec
i
a
ll
y
u
s
ed
t
o repre
s
en
t
da
t
a
i
n a
m
a
t
r
i
x for
m
.
Dec
l
ara
ti
on of
t
wo d
im
en
si
ona
l
array
s:
Syn
t
ax:-
Da
t
a
t
ype arrayna
m
e [row
si
ze] [co
l
u
m
n
si
ze]
;
Eg:
i
n
t
arr[3][4]
;
and 4 co
l
u
m
n
s
.
A
t
wo-d
im
en
si
ona
l
array can be con
si
dered a
s
a
t
ab
l
e wh
i
ch w
ill
have
x nu
m
ber of row
s
and y nu
m
ber of co
l
u
m
n
s
. A
t
wo-d
im
en
si
ona
l
array
a, wh
i
ch con
t
a
i
n
s
t
hree row
s
and four co
l
u
m
n
s
can be
s
hown a
s
fo
ll
ow
s
4M
E
xp
l
a
n
a
ti
o
n
2
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
Thu
s
, every e
l
e
m
en
t
i
n
t
he array a
is
i
den
ti
f
i
ed by an e
l
e
m
en
t
na
m
e of
t
he for
m
a[
i
][
j
], where 'a'
is
t
he na
m
e of
t
he array, and '
i
' and '
j
' are
t
he
s
ub
s
cr
i
p
ts
t
ha
t
un
i
que
l
y
i
den
ti
fy each e
l
e
m
en
t
i
n 'a'.
E
xam
pl
e :
m
a
i
n()
{
i
n
t
a[2][2]
=
{{1,2},{4,5})
;
i
n
t
i
,
j;
for(
i=
0
;i<
2
;i++
)
{
for(
j=
0
;j<
2
;j++
)
{
}
\
}
}
E
xa
m
p
l
e
2
M
(
d
)
A
ns
.
E
x
pl
a
in
a
n
y two
s
tr
in
g f
un
ct
i
o
ns
w
i
t
h
exam
pl
e.
S
tr
l
e
n
f
un
ct
i
o
n
:
st
r
l
en( ) func
ti
on
i
n
C
g
i
ve
s
t
he
l
eng
t
h of
t
he g
i
ven
st
r
i
ng.
st
r
l
en( )
func
ti
on coun
ts
t
he nu
m
ber of charac
t
er
s
i
n a g
i
ven
st
r
i
ng and re
t
urn
s
t
he
i
n
t
eger va
l
ue. I
t
st
op
s
coun
ti
ng
t
he charac
t
er when nu
ll
charac
t
er
is
found.
B
ecau
s
e, nu
ll
charac
t
er
i
nd
i
ca
t
e
s
t
he end of
t
he
st
r
i
ng
i
n
C
.
S
y
nt
ax:
s
tr
l
e
n
(
s
tr
in
g
n
ame);
E
xa
m
p
l
e:
st
r
l
en(
st
r1)
;
re
t
urn
s
l
eng
t
h of
st
r1 a
s
3
4M
E
xp
l
a
n
a
ti
o
n
of
a
n
y
tw
o
strin
g
f
un
c
ti
o
n
s
1
M
eac
h
,
exa
m
p
l
e
1
M
eac
h
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
s
trcat() f
un
ct
i
o
n
:
In
C
progra
mmi
ng,
st
rca
t
() conca
t
ena
t
e
s
(
j
o
i
n
s
)
t
wo
st
r
i
ng
s
. I
t
conca
t
ena
t
e
s
s
ource
st
r
i
ng a
t
t
he end of de
sti
na
ti
on
st
r
i
ng.
S
y
nt
ax:
s
trcat(
d
e
s
t
in
at
i
o
ns
o
u
rce,
s
o
u
rce
s
tr
in
g);
E
xa
m
p
l
e:
st
rca
t
(
st
r1,
st
r2)
;
re
t
urn
s
abcdef
i
n
st
r1 and
st
r2 re
m
a
i
n
s
unchanged.
s
trc
p
y() f
un
ct
i
o
n
st
rncpy( ) func
ti
on cop
i
e
s
por
ti
on of con
t
en
ts
of one
st
r
i
ng
i
n
t
o
ano
t
her
st
r
i
ng.
S
y
nt
ax:
st
rncpy( de
sti
na
ti
on
st
r
i
ng,
s
ource
st
r
i
ng,
si
ze )
;
E
xa
m
p
l
e:
st
rcpy(
st
r1,
st
r2)
;
re
t
urn
s
abc
st
r2
s
trcm
p
() f
un
ct
i
o
n
The
st
rc
m
p func
ti
on co
m
pare
s
t
wo
st
r
i
ng
s
wh
i
ch are pa
ss
ed a
s
argu
m
en
ts
t
o
it
. If
t
he
st
r
i
ng
s
are equa
l
t
hen func
ti
on re
t
urn
s
va
l
ue 0 and
i
f
t
hey are no
t
equa
l
t
he func
ti
on
re
t
urn
s
s
o
m
e nu
m
er
i
c va
l
ue.
S
y
nt
ax:
st
rc
m
p(
st
r1,
st
r2)
;
E
xa
m
p
l
e:
Then
st
rc
m
p(
st
r1,
st
r2) re
t
urn
s
0 a
s
bo
t
h
t
he
st
r
i
ng
s
are
s
a
m
e.
4.
(a)
A
ns
.
Attem
p
t a
n
y
TH
R
EE
of t
h
e fo
ll
ow
in
g:
Draw f
l
owc
h
art for f
indin
g
l
arge
s
t
nu
m
b
er amo
n
g t
h
ree
nu
m
b
er
s
.
12
4M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
C
o
rr
ec
t
f
l
o
w
c
h
a
r
t
4
M
(
b
)
A
ns
.
4M
L
ist
of
s
ec
ti
o
ns
f
r
o
m
stru
c
tur
e 1
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
Doc
u
me
n
tat
i
o
n
s
ect
i
o
n
:
The docu
m
en
t
a
ti
on
s
ec
ti
on con
sists
of a
s
e
t
of co
mm
en
t
li
ne
s
g
i
v
i
ng
t
he na
m
e of
t
he progra
m
,
t
he au
t
hor and
o
t
her de
t
a
ils
, wh
i
ch
t
he progra
mm
er wou
l
d
li
ke
t
o u
s
e
l
a
t
er.
Link
s
ect
i
o
n
:
The
li
nk
s
ec
ti
on prov
i
de
s
i
n
st
ruc
ti
on
s
t
o
t
he co
m
p
il
er
t
o
li
nk func
ti
on
s
fro
m
t
he
s
y
st
e
m
li
brary
s
uch a
s
u
si
ng
t
he #
i
nc
l
ude
d
i
rec
ti
ve.
Def
ini
t
i
o
n
s
ect
i
o
n
:
The def
i
n
iti
on
s
ec
ti
on def
i
ne
s
a
ll
s
y
m
bo
li
c
con
st
an
ts
s
uch u
si
ng
t
he #def
i
ne d
i
rec
ti
ve.
Gl
o
b
a
l
d
ec
l
arat
i
o
n
s
ect
i
o
n
:
There are
s
o
m
e var
i
ab
l
e
s
t
ha
t
are u
s
ed
i
n
m
ore
t
han one func
ti
on.
S
uch var
i
ab
l
e
s
are ca
ll
ed g
l
oba
l
var
i
ab
l
e
s
and
are dec
l
ared
i
n
t
he g
l
oba
l
dec
l
ara
ti
on
s
ec
ti
on
t
ha
t
is
ou
tsi
de of a
ll
t
he
func
ti
on
s
.
Dec
l
arat
i
o
n
p
art
:
The dec
l
ara
ti
on par
t
dec
l
are
s
a
ll
t
he var
i
ab
l
e
s
u
s
ed
i
n
t
he execu
t
ab
l
e par
t
.
Subp
rogram
s
ect
i
o
n
:
If
t
he progra
m
is
a
m
u
lti
-func
ti
on progra
m
t
hen
t
he
s
ubprogra
m
s
ec
ti
on con
t
a
i
n
s
a
ll
t
he u
s
er-def
i
ned func
ti
on
s
t
ha
t
are ca
ll
ed
i
n
t
he
m
a
i
n () func
ti
on. U
s
er-def
i
ned func
ti
on
s
are
genera
ll
y p
l
aced
imm
ed
i
a
t
e
l
y af
t
er
t
he
m
a
i
n () func
ti
on, a
lt
hough
t
hey
m
ay appear
i
n any order.
H
ea
d
er f
il
e
s
A header f
il
e
is
a f
il
e w
it
h ex
t
en
si
on .h wh
i
ch con
t
a
i
n
s
C
func
ti
on
dec
l
ara
ti
on
s
and
m
acro def
i
n
iti
on
s
t
o be
s
hared be
t
ween
s
evera
l
s
ource f
il
e
s
.
In
c
lud
e
S
y
n
tax
B
o
t
h
t
he u
s
er and
t
he
s
y
st
e
m
header f
il
e
s
are
i
nc
l
uded u
si
ng
t
he
preproce
ssi
ng d
i
rec
ti
ve #
i
nc
l
ude.
ma
in
() func
ti
on
is
t
he en
t
ry po
i
n
t
of any
C
progra
m
. I
t
is
t
he po
i
n
t
a
t
wh
i
ch execu
ti
on of progra
m
is
st
ar
t
ed. Every
C
progra
m
have a
m
a
i
n() func
ti
on.
C
o
rr
ec
t
de
s
c
ri
p
ti
o
n
of
stru
c
tur
e 3
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
(c)
A
ns
.
Wr
i
te a
p
rogram to ta
k
e
inpu
t a
s
a
nu
m
b
er a
nd
rever
s
e
i
t
b
y
w
hil
e
l
oo
p
.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
i
n
t
no
;
i
n
t
s
u
m=
0,re
m;
pr
i
n
t
f("
\
n En
t
er nu
m
ber
:
")
;
s
canf("%d",
&
no)
;
wh
il
e(no
>
0)
{
re
m=
no%10
;
no
=
no
/
10
;
s
u
m=s
u
m
*10
+
r e
m;
}
pr
i
n
t
f("
\
n
s
u
m=
%d",
s
u
m
)
;
ge
t
ch()
;
}
4M
A
ccep
t
in
p
ut
1
M
U
s
e of
whil
e
l
oop 1
M
co
rr
ec
t
s
y
nt
ax
2
M
(
d
)
A
ns
.
Wr
i
te a
p
rogram to acce
p
t 10
nu
m
b
er
s
in
array a
nd
arra
n
ge
t
h
em
in
a
s
ce
ndin
g or
d
er.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
i
n
t
arr[10],
i
,
j
,
t
e
m
p
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er array e
l
e
m
en
ts:
")
;
for(
i=
0
;i<
10
;i++
)
{
s
canf("%d",
&
arr[
i
])
;
}
pr
i
n
t
f("
\
n
\
n Array e
l
e
m
en
ts
are
:
")
;
for(
i=
0
;i<
10
;i++
)
{
pr
i
n
t
f("%d ",arr[
i
])
;
4M
C
o
rr
ec
t
l
og
i
c
2
M
C
o
rr
ec
t
s
y
nt
ax
2
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
}
for(
j=
0
;j<
10
;j++
)
{
for(
i=
0
;i<
10
;i++
)
{
i
f(arr[
i+
1]
<
arr[
i
])
{
t
e
m
p
=
arr[
i
]
;
arr[
i
]
=
arr[
i+
1]
;
arr[
i+
1]
=t
e
m
p
;
}
}
}
pr
i
n
t
f("
\
n
\
nArray e
l
e
m
en
ts
i
n a
s
cend
i
ng order are
:
")
;
for(
i=
0
;i<
10
;i++
)
{
pr
i
n
t
f("%d ",arr[
i
])
;
}
ge
t
ch()
;
}
(e)
A
ns
.
E
x
pl
a
in
mea
nin
g of fo
ll
ow
in
g
s
tateme
n
t w
i
t
h
refere
n
ce to
p
o
in
ter
s
:
in
t *a,
b
;
b=
20;
* a
=b
;
A
=
&
b
;
in
t *a,
b
;
I
t
is
dec
l
ara
ti
on of
i
n
t
eger po
i
n
t
er a and
i
n
t
eger var
i
ab
l
e b
b=
20;
va
l
ue 20
is
a
ssi
gned
t
o var
i
ab
l
e b.
*a
=b
;
Va
l
ue of b
is
a
ssi
gned
t
o po
i
n
t
er a.
A
=
&
b
;
Addre
ss
of b
is
a
ssi
gned
t
o var
i
ab
l
e A.
4M
C
o
rr
ec
t
m
ea
nin
g
of eac
h
st
a
t
e
m
e
n
t
1
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
5.
(a)
A
ns
.
Attem
p
t a
n
y
T
W
O
of t
h
e fo
ll
ow
in
g:
Wr
i
te a
p
rogram to
p
erform a
ddi
t
i
o
n
,
sub
tract
i
o
n
, m
ul
t
ipli
cat
i
o
n
a
nd
di
v
isi
o
n
of two
in
teger
nu
m
b
er
usin
g f
un
ct
i
o
n
.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d add(
i
n
t
x,
i
n
t
y)
{
pr
i
n
t
f("
\
nAdd
iti
on
=
%d",x
+
y)
;
}
vo
i
d
s
ub(
i
n
t
x,
i
n
t
y)
{
pr
i
n
t
f("
\
n
S
ub
t
rac
ti
on
=
%d",x-y)
;
}
vo
i
d
m
u
lt
(
i
n
t
x,
i
n
t
y)
{
pr
i
n
t
f("
\
n
M
u
lti
p
li
ca
ti
on
=
%d",x*y)
;
}
vo
i
d d
i
v(
i
n
t
x,
i
n
t
y)
{
pr
i
n
t
f("
\
nD
i
v
isi
on
=
%d",x
/
y)
;
}
vo
i
d
m
a
i
n()
{
i
n
t
x,y
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er x
:
")
;
s
canf("%d",
&
x)
;
pr
i
n
t
f("En
t
er y
:
")
;
s
canf("%d",
&
y)
;
add(x,y)
;
s
ub(x,y)
;
m
u
lt
(x,y)
;
d
i
v(x,y)
;
ge
t
ch()
;
}
12
6M
A
dd
f
un
c
ti
o
n
1
M
su
b
f
un
c
ti
o
n
1
M
Mult
f
un
c
ti
o
n
1
M
D
i
v
f
un
c
ti
o
n
1
M
M
a
in
f
un
c
ti
o
n
2
M
(
b
)
Def
in
e Array. Wr
i
te a
p
rogram to acce
p
t te
n
nu
m
b
er
s
in
array.
S
ort array e
l
eme
n
t a
nd
displ
ay.
6M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
A
ns
.
Def
ini
t
i
o
n
of Array:
An array
is
a co
ll
ec
ti
on of da
t
a e
l
e
m
en
ts
, a
ll
of
t
he
s
a
m
e
t
ype,
acce
ss
ed u
si
ng a co
mm
on na
m
e.
Program:
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
i
n
t
a[10],
i
,
j
,
t
e
m
p
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er nu
m
ber
s:
")
;
for(
i=
0
;i<
10
;i++
)
s
canf("%d",
&
a[
i
])
;
for(
i=
0
;i<
10
;i++
)
{
for(
j=i+
1
;j<
10
;j++
)
{
i
f(a[
i
]
>
a[
j
])
{
t
e
m
p
=
a[
i
]
;
a[
i
]
=
a[
j
]
;
a[
j
]
=t
e
m
p
;
}
}
}
pr
i
n
t
f("
\
n
S
or
t
ed array e
l
e
m
en
ts:
")
;
for(
i=
0
;i<
10
;i++
)
pr
i
n
t
f("
\
n %d",a[
i
])
;
ge
t
ch()
;
}
Arr
ay
def
initi
o
n
1
M
A
ccep
tin
g a
rr
ay
1
M
S
o
rtin
g
l
og
i
c 3
M
D
is
p
l
ay
s
o
rt
ed
a
rr
ay
1
M
(c)
A
ns
.
Wr
i
te a
p
rogram to
p
r
in
t rever
s
e of a e
n
tere
d
s
tr
in
g
usin
g
p
o
in
ter.
(No
t
e:
An
y o
th
e
r
co
rr
ec
t
l
og
i
c
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
4M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
char
st
r[10],*p
t
r
;
i
n
t
l=
0
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er
st
r
i
ng
:
")
;
s
canf("%
s
",
st
r)
;
p
t
r
=st
r
;
wh
il
e(*p
t
r!
=
'
\
0')
{
l=l+
1
;
p
t
r
=
p
t
r
+
1
;
}
wh
il
e(
l>
0)
{
p
t
r
=
p
t
r-1
;
pr
i
n
t
f("%c",*p
t
r)
;
l=l
-1
;
}
ge
t
ch()
;
}
A
ccep
tin
g
strin
g
1
M
po
int
e
r
initi
a
liz
a
ti
o
n
1
M
l
og
i
c of
r
eve
rs
e
usin
g
po
int
e
r
3
M
D
is
p
l
ay
i
n
g
r
eve
rs
e
strin
g
1
M
6.
(a)
A
ns
.
Attem
p
t a
n
y
T
W
O
of t
h
e fo
ll
ow
in
g:
E
x
pl
a
in
rec
u
r
si
o
n
w
i
t
h
sui
ta
bl
e exam
pl
e.
Lis
t a
n
y two
a
d
va
n
tage
s
.
R
ecur
si
on
m
ean
s
a func
ti
on ca
lls
its
e
l
f repe
titi
ve
l
y. A recur
si
ve
func
ti
on con
t
a
i
n
s
a func
ti
on ca
ll
t
o
its
e
l
f
i
n
si
de
its
body.
E
xa
m
p
l
e:
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
i
n
t
fac
t
or
i
a
l
(
i
n
t
N)
;
vo
i
d
m
a
i
n()
{
i
n
t
N,fac
t;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er nu
m
ber
:
")
;
s
canf("%d",
&
N)
;
fac
t=
fac
t
or
i
a
l
(N)
;
12
6M
E
xp
l
a
n
a
ti
o
n
of
r
ec
ursi
o
n
1
M
E
xa
m
p
l
e
3
M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
pr
i
n
t
f("
\
n
F
ac
t
or
i
a
l
is:
%d",fac
t
)
;
ge
t
ch()
;
}
i
n
t
fac
t
or
i
a
l
(
i
n
t
N)
{
i
f(N
==
1)
re
t
urn(1)
;
e
ls
e
re
t
urn(N*fac
t
or
i
a
l
(N-1))
;
}
A
d
va
n
tage
s
:
R
educe
s
l
eng
t
h of
t
he progra
m
R
educe
s
unnece
ss
ary ca
lli
ng of a func
ti
on.
U
s
efu
l
when
s
a
m
e
s
o
l
u
ti
on
is
t
o be app
li
ed
m
any
tim
e
s
.
An
y
tw
o
A
dva
nt
a
ge
s
2
M
(
b
)
A
ns
.
Wr
i
te a
p
rogram to acce
p
t te
n
nu
m
b
er
s
a
nd
p
r
in
t average of
i
t.
(No
t
e: P
r
og
r
a
m
with
o
ut
a
rr
ay
sh
a
ll
be co
nsi
de
r
ed).
#
i
nc
l
ude
<st
d
i
o.h
>
#
i
nc
l
ude
<
con
i
o.h
>
vo
i
d
m
a
i
n()
{
i
n
t
a[10],
i
,
s
u
m=
0
;
f
l
oa
t
avg
;
c
l
r
s
cr()
;
pr
i
n
t
f("En
t
er nu
m
ber
s:
")
;
for(
i=
0
;i<
10
;i++
)
s
canf("%d",
&
a[
i
])
;
for(
i=
0
;i<
10
;i++
)
s
u
m=s
u
m+
a[
i
]
;
avg
=s
u
m/
10
;
pr
i
n
t
f("
\
n Average
=
%f", avg)
;
ge
t
ch()
;
}
6M
A
ccep
tin
g 10
num
be
rs
2
M
C
a
l
c
ul
a
t
in
g
ave
r
age
2
M
D
is
p
l
ay
i
n
g
ave
r
age
2
M
(c)
A
ns
.
Enlis
t
di
ffere
n
t format
sp
ec
i
f
i
er
s
w
i
t
h
i
t
s
us
e.
F
or
m
a
t
s
pec
i
f
i
er
t
e
lls
t
he co
m
p
il
er wha
t
t
ype of da
t
a a var
i
ab
l
e ho
l
d
s
dur
i
ng
t
ak
i
ng
i
npu
t
and pr
i
n
ti
ng ou
t
pu
t
u
si
ng
s
canf() and pr
i
n
t
f()
func
ti
on
s
re
s
pec
ti
ve
l
y.
F
or
m
a
t
s
pec
i
f
i
er
s
u
s
ed
i
n
C
progra
mmi
ng
:
6M
MA
H
ARA
SHT
RA
ST
A
TE
BO
ARD
O
F
TE
C
H
N
I
CA
L
E
DUCA
TIO
N
(A
u
to
n
omo
us
)
(
ISO/IE
C - 27001 - 2005 Cert
i
f
i
e
d
)
S
UMM
E
R 2019
E
XAM
I
NA
TIO
N
M
O
D
EL
AN
S
W
E
R
Sub
ject: Programm
in
g
in
C
Sub
ject Co
d
e:
22226
F
or
m
a
t
s
pec
i
f
i
er
U
s
e
%d
S
pec
i
fy da
t
a
t
ype a
s
s
hor
t
si
gned
%u
S
pec
i
fy da
t
a
t
ype a
s
s
hor
t
un
si
gned
%
l
d
S
pec
i
fy da
t
a
t
ype a
s
l
ong
si
nged
%
l
u
S
pec
i
fy da
t
a
t
ype a
s
l
ong un
si
gned
%x
S
pec
i
fy da
t
a
t
ype a
s
un
si
gned hexadec
im
a
l
%o
S
pec
i
fy da
t
a
t
ype a
s
un
si
gned oc
t
a
l
%f
S
pec
i
fy da
t
a
t
ype a
s
f
l
oa
t
%
l
f
S
pec
i
fy da
t
a
t
ype a
s
doub
l
e
%Lf
S
pec
i
fy da
t
a
t
ype a
s
l
ong doub
l
e
%c
S
pec
i
fy da
t
a
t
ype a
s
si
gned charac
t
er
%
s
S
pec
i
fy da
t
a
t
ype a
s
un
si
gned group of
charac
t
er
s
(
St
r
i
ng
s
)
An
y
si
x
fo
rm
a
t
s
pec
i
f
i
e
r
s
with
us
e 1
M
eac
h
22226
[1 of 2] P.T.O.
11920
3 Hours / 70 Marks Seat No.
Instructions : (1) All Questions are compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Mobile Phone, Pager and any other Electronic Communication
devices are not permissible in Examination Hall.
Marks
1. Attempt any FIVE of the following : 10
(a) Define array. List its type.
(b) Draw & label different symbols used in flowcharts.
(c) Find the output of the following program :
# include < stdio.h>
void main( )
{
int x = 10, y = 10, v1, v2 ;
v1 = x++ ;
v2 = ++y ;
printf ("value of v1:%d", v1) ;
printf ("value of v2:%d", v2) ;
}
(d) State the syntax & use of strlen ( ) & strcat ( ) function.
(e) State the Relational operators with example.
(f) State the syntax to declare pointer variable with example.
(g) Draw flow chart for addition of two numbers.
22226 [2 of 2]
2. Attempt any THREE of the following : 12
(a) State the importance of flow chart.
(b) Write a program to declare structure student having rollno, name & marks.
Accept & display data for 3 students.
(c) Explain pointer arithmetic with example.
(d) Explain nested if-else with example.
3. Attempt any THREE of the following : 12
(a) Describe the following terms :
(i) Keyword
(ii) Identifier
(iii) Variable
(iv) Constant
(b) Differentiate between call by value and call by reference.
(c) Explain conditional operator with example.
(d) List the categories of functions and explain any one with example.
4. Attempt any THREE of the following : 12
(a) Write an algorithm to determine the given number is odd or even.
(b) Illustrate the use of break and continue statement with example.
(c) Write a program to add, subtract, multiply and divide two numbers, accepted
from user using switch case.
(d) Illustrate initialization of two dimensional array with example.
(e) Write a program to read two strings and find whether they are equal or not.
5. Attempt any TWO of the following : 12
(a) Write a program to calculate sum of all the odd numbers between 1 to 20.
(b) Write a program for addition of two 3 3 matrices.
(c) Write a program to compute the sum of all elements stored in an array using
pointers.
6. Attempt any TWO of the following : 12
(a) Write a program to sort elements of an array in ascending order.
(b) Write a function to print Fibonacci series starting from 0, 1.
(c) Calculate factorial of a number using recursion.
_______________
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
WINTER – 2019 EXAMINATION
MODEL ANSWER
Subject: Programming in 'C' Subject Code:
Page 1 / 17
22226
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance
(Not applicable for subject English and Communication Skills).
4) While assessing figures, examiner may give credit for principal components indicated in the
figure. The figures drawn by candidate and model answer may vary. The examiner may give
credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model
answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant
answer based on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
Q.
No
.
Sub
Q.N.
Answer
Marking
Scheme
1.
(a)
Ans.
Attempt any FIVE of the following:
Define array. List its type.
Array is a fixed-size sequential collection of elements of the same
type.
Types:
1. One dimensional
2. Multi dimensional
10
2M
Definitio
n 1M
Types
1M
(b)
Ans.
Draw & label different symbols used in flowcharts.
Symbol
Name
Function
Start/end
An oval represents a start or end
point
Arrows
A line is a connector that shows
relationships between the
representative shapes
2M
Any 4
symbols
½M
each
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
WINTER – 2019 EXAMINATION
MODEL ANSWER
Subject: Programming in 'C' Subject Code:
Page 2 / 17
22226
Input/Output
A parallelogram represents input
or output
Process
A rectangle represents a process
Decision
A diamond indicates a decision
(c)
Ans.
Find the output of the following program:
#include<stdio.h>
void main( )
{
int x = 10, y = 10, v1, v2;
v1 = x++;
v2 = ++y;
printf(“value of v1: %d, v1);
printf(“value of v2: %d, v2);
}
Output:
value of v1:10value of v2:11
2M
Correct
output
2M
(d)
Ans.
State the syntax & use of strlen ( ) & strcat ( ) function.
strlen( ): calculates the length of the string
Syntax: strlen(s1);
strcat():concatenates two strings
Syntax: strcat(s1,s2)
2M
1M for
correct
syntax
1M for
use
(e)
Ans.
State the Relational operators with example.
== - returns true if the values of two operands are equal else returns
false.
E.g: if (A= = B){ }
!= - returns true if values of two operands are not equal, else returns
2M
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
WINTER – 2019 EXAMINATION
MODEL ANSWER
Subject: Programming in 'C' Subject Code:
Page 3 / 17
22226
false
E.g: if (A! = B){ }
<- returns true if the first operand is less than the second, else returns
false.
E.g: if (A< B){ }
>- returns true if the first operand is greater than the second, else
returns false.
E.g: if (A> B){ }
<= returns true if the first operand is less than or equal to the second,
else returns false.
E.g: if (A< = B){ }
>= returns true if the first operand is greater than or equal to the
second, else returns false.
E.g: if (A> = B){ }
Any
four
operator
s ½M
each
(f)
Ans.
State the syntax to declare pointer variable with example.
General syntax to declare pointer.
datatype *var_name;
Eg: int var = 20;
2M
Correct
syntax
1M
Correct
example
1M
(g)
Ans.
Draw flow chart for addition of two numbers.
2M
Correct
sequenc
e 1M
Correct
symbol
1M
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
WINTER – 2019 EXAMINATION
MODEL ANSWER
Subject: Programming in 'C' Subject Code:
Page 4 / 17
22226
2.
(a)
Ans.
Attempt any THREE of the following:
State the importance of flow chart.
A flowchart is a type of diagram that represents an algorithm. It is a
visual representation of a sequence of steps to complete the process.
A flow chart describes a process using symbols rather than words.
Computer programmers use flow charts to show where data enters the
program, what processes the data goes through, and how the data is
converted to output.
-can be used to quickly communicate the ideas or plans that one
programmer envisions to other people who will be involved in the
process.
- aid in the analysis of the process to make sure nothing is left out and
that all possible inputs, processes, and outputs have been accounted
for.
-help programmers develop the most efficient coding because they
can clearly see where the data is going to end up.
- help programmers figure out where a potential problem area is and
helps them with debugging or cleaning up code that is not working.
- are a useful tool in visualizing a module's flow of execution before
writing any code. This allows developers to do three things: verify the
algorithm's correctness before writing code, visualize how the code
will ultimately be written, and communicate and document the
algorithm with other developers and even non-developers.
-may be used in conjunction with other tools, such as pseudo-code, or
may be used by itself to communicate a module's ultimate design,
depending on the level of detail of the flowchart.
12
4M
Any 4
points
1M each
(b)
Ans.
Write a program to declare structure student having rollno,
name & marks.
(Note: Any other correct logic shall be considered).
Accept and display data for three students.
#include<stdio.h>
#include<conio.h>
void main() {
int i;
struct student{
int rollno;
char name[20];
int marks;
} s[3];
4M
Correct
logic 3M
Correct
syntax
1M
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clrscr();
for(i=0;i<3;i++) {
printf("Enter rollno, name and marks\n");
scanf("%d%s%d",&s[i].rollno,&s[i].name,&s[i].marks);
}
for(i = 0; i<3;i++){
printf("\nThe details of student %d\n",i+1);
printf("Roll no %d\n",s[i].rollno);
printf("Name is %s\n",s[i].name);
printf("Marks %d\n",s[i].marks);
}
getch();
}
(c)
Ans.
Explain pointer arithmetic with example.
(Note: Code snippet shall be considered).
The pointer arithmetic is done as per the data type of the pointer. The
basic operations on pointers are:
Increment
It is used to increment the pointer. Each time a pointer is
incremented, it points to the next location. Eg, for an int pointer
variable, if the current position of pointer is 1000, when incremented
it points to 1002 because for storing an int value it takes 2 bytes of
memory.
Decrement
It is used to decrement the pointer. Each time a pointer is
decremented, it points to the previous location. Eg, if the current
position of pointer is 1002, then decrement operation results in the
pointer pointing to the location 1000.
Addition and subtraction:
When addition or subtraction operation is performed on the pointer
variable, it shows that particular location in the memory.
Eg: int *ptr; -say address is 1000.
If -> ptr+n- then ptr+n*2 .
If -> ptr-n thenptr-n*2.
#include<stdio.h>
#include<conio.h>
void main() {
4M
Any two
operator
s
Each
operator
with
explanat
ion 1M
1M for
each
example
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int i = 10;
int *ptr=&i;
clrscr();
printf("%x%d",ptr,i);
ptr++;
printf("\n%x%d",ptr,i);
printf("\n%x",ptr+2);
printf("\n%x",ptr-2);
getch();
}
(d)
Ans.
Explain nested if-else with example.
(Note: Any example shall be considered)
When a series of decision is required, nested if-else is used. Nesting
means using one if-else construct within another one. If the condition
in the outer if, is true, then only the inner if-else will get executed.
Further the statements in the inner if will get execute only if the
condition of inner if, evaluates to true. If it is false, the statements in
inner else will get executed.
If the outer if evaluates to false, then the statements in outer else get
executed.
General syntax:
if(condition) {
if(condition) {
statements
} else {
statements
}
} else {
statements
}
statements
Example:
#include<stdio.h>
#include<conio.h>
void main() {
int val;
clrscr();
4M
Explana
tion 2M
Example
2M
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printf("Enter a number");
scanf("%d",&val);
if(val>=5) {
if(val>5) {
printf("Number is greater than 5");
} else {
printf("Number is equal to 5");
}
} else {
printf("Number is less than 5");
}
getch();
}
3.
(a)
Ans.
Attempt any THREE of the following:
Describe the following terms:
(i) Keyword
(ii) Identifier
(iii) Variable
(iv) Constant
(i) Keyword: Keywords are special words in C programming which
have their own predefined meaning. The functions and meanings of
these words cannot be altered. Some keywords in C Programming
are if, while, for, do, etc..
(ii) Identifier: Identifiers are user-defined names of variables,
functions and arrays. It comprises of combination of letters and digits.
Example
int age1;
float height_in_feet;
Here, age1 is an identifier of integer data type.
Similarly height_feet is also an identifier but of floating integer data
type,
(iii) Variable: A variable is nothing but a name given to a storage
area that our programs can manipulate. Each variable in C has a
specific type, which determines the size and layout of the variable's
memory; the range of values that can be stored within that memory;
and the set of operations that can be applied to the variable.
Example: add, a, name
(iv) Constant:
12
4M
Each
term 1M
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Constants refer to fixed values that the program may not change
during its execution. These fixed values are also called literals.
Constants can be of any of the basic data types like an integer
constant, a floating constant, a character constant, or a string literal.
There are enumeration constants as well.
Example:
121
234
3.14
(b)
Ans.
Differentiate between call by value and call by reference.
Sr.
No.
Call by value
Call by reference
1
When function is called
by passing values then it
is call by value
When function is called by
passing address of variable then
it is called as call by reference.
2
Copy of actual variable is
created when function is
called.
No copy is generated for actual
variable rather address of actual
variable is passed.
3
In call by value, memory
required is more as copy
of variable is created.
In call by reference, memory
required is less as there is no
copy of actual variables.
4
Example:-
Function call -
Swap ( x,y);
Calling swap function by
passing
values.
Example:-
Function call –
Swap ( &x, &y );
Calling swap function by
passing
address.
5
Original (actual)
parameters do not change.
Changes take place on the
copy of variable.
Actual parameters change as
function operates on value
stored at the address.
4M
Any
four
differen
ces 1M
each
(c)
Ans.
Explain conditional operator with example.
Conditional Operator (Ternary Operator):
It takes the form „? :‟ to construct conditional expressions
The operator „? :‟ works as follows:
exp1 ? exp2 : exp 3
Where exp1, exp2 and exp3 are expressions.exp1 is evaluated first, If
it is true, then the expression exp2 is evaluated and becomes the value
4M
Explana
tion 2M
Example
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of the expression. If exp1 is false, exp3 is evaluated and its value
becomes the value of the expression.
E.g. int a=10,b=5,x;
x=(a>b) ? a : b;
2M
(d)
Ans.
List the categories of functions and explain any one with example.
Different categories of function:
1) Function with no arguments and no return value.
2) Function with arguments and no return value.
3) Function with no arguments and return value.
4) Function with arguments and return value.
1) Function with no arguments and no return value:
This category of function cannot return any value back to the calling
program and it does not accept any arguments also. It has to be
declared as void.
For example:
void add()
{
inta,b,c;
a=5;
b=6;
c=a+b;
printf(“%d”,c);
}
It should be called as add();
2) Function with arguments and no return value:
This category of function cannot return any value back to the calling
program but it takes arguments from calling program. It has to be
declared as void. The number of arguments should match in
sequence, number and data type.
For example:
void add(intx,int y)
{
int z;
z=x+y;
printf(“%d”,z);
}
It should be called as add(4,5); where x will take 4 and y will take 5
as their values.
4M
List 2M
Explana
tion of
any one
category
2M
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3) Function with no arguments and return value:
This category of function can return a value back to the calling
program but it does not take arguments from calling program. It has
to be declared with same data type as the data type of return variable.
For example:
int add()
{
inta,b,c;
a=5;
b=6;
c=a+b;
return(c);
}
It should be called as int x = add(); where x will store value returned
by the function.
4) Function with arguments and return value:
This category of function can return a value back to the calling
program but it also takes arguments from calling program. It has to be
declared with same data type as the data type of return variable.
For example:
int add(intx,int y)
{
int z;
z=x+y;
return(z);
}
It should be called as int s = add(4,5); where x will have 4 and y will
have 5 as their values and s will store value returned by the function.
4.
(a)
Ans.
Attempt any THREE of the following:
Write an algorithm to determine the given number is odd or
even.
Step 1- Start
Step 2- Read / input the number.
Step 3- if n%2==0 then number is even.
Step 4- else number is odd.
Step 5- display the output.
Step 6- Stop
12
4M
Correct
algorith
m 4M
(b)
Illustrate the use of break and continue statement with example.
4M
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Ans.
(Note:- Any other example shall be considered)
Break: It breaks the execution of the loop which allows exiting from
any loop or switch, such that break statement skips the remaining part
of current iterations of the loop.
Syntax: break;
Continue: It is used when it is required to skip the remaining portion
of the loop without breaking loop it will transfer control directly to
next iteration
Syntax: continue;
In given program sequence if “break” executes then execution control
will jump out of loop & next statement after loop will be executed. In
given program sequence if “continue” executes then execution
control will skip remaining statements of loop & will start next
iteration of loop
Use of
each 1M
Example
of each
1M
(c)
Ans.
Write a program to add, subtract, multiply and divide two
numbers, accepted from user switch case.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b,ch,add,sub,mul,div;
clrscr();
printf("\n1 for addition \n2 for substraction");
printf("\n3 for multiplication \n4 for division");
printf("\nEnter two numbers:");
4M
Correct
logic 2M
Correct
syntax
2M
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scanf("%d%d",&a,&b);
printf("\nEnter your choice:");
scanf("%d",&ch);
switch(ch)
{
case 1:
add=a+b;
printf("Addition of a & b=%d",add);
break;
case 2:
sub=a-b;
printf("Substraction of a & b=%d",sub);
break;
case 3:
mul=a*b;
printf("Multiplication of two numbers=%d",mul);
break;
case 4:
div=a/b;
printf("Division of two numbers=%d",div);
break;
default:
printf("Invalid choice....");
}
getch();
}
(d)
Ans.
Illustrate initialization of two dimensional array with example.
Two dimensional array:
The array which is used to represent and store data in a tabular form
is called as two dimensional array. Such type of array is specially
used to represent data in a matrix form.
Initialization can be done as design time or runtime.
1. Design time: This can be done by providing „row X column‟
number of elements to the array. Eg for a 3 rows and 4 columns array
, 3X4=12 elements can be provided as :
arr[3][4]={ {2,3,4,6},
{1,4,6,3},
{6,6,4,3},
{6,7,8,9}
};
4M
Two dim
array
1M
Declarat
ion 1M
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2. Runtime: For this loop structures like „for‟ can be used in a nested
form, where outer loop will increment row and inner loop will
increment column.
Eg :
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
scanf(“%d”, &arr[i][j]);
}
}
Example:
main()
{
int arr[2][2]={{1,2},{4,5});
int i,j;
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
printf( “%d”, arr[i][j]);
}
printf(“\n”);
}
}
Initializ
ation by
any one
type 1M
Example
1M
(e)
Ans.
Write a program to read two strings and find whether they are
equal or not.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
char st1[20],st2[20];
printf(“enter string 1”);
scanf(“%s”,st1);
printf(“enter second string”);
scanf(“%s”,st2);
if(strcmp(st1,st2)==0)
4M
Correct
logic 2M
Correct
syntax
2M
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printf(“\nboth strings are equal”);
else
printf(“\nstrings are not equal”);
}
5.
(a)
Ans.
Attempt any TWO of the following:
Write a program to calculate sum of all the odd numbers between
1 to 20.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
void main()
{
inti,sum=0;
clrscr();
for(i=1;i<=20;i++)
{
if(i%2!=0)
{
sum=sum+i;
}
}
printf("Sum=%d",sum);
getch();
}
12
6M
Finding
odd
numbers
2M
Calculat
ing sum
1M
Display
sum 1M
Correct
syntax
2M
(b)
Ans.
Write a program for addition of two 3 x 3 matrices.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
void main()
{
int a[3][3],b[3][3],c[3][3],i,j;
clrscr();
printf("\n Enter first matrix");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
scanf("%d",&a[i][j]);
}
6M
Decelera
tion of
variable
s 1M
Input
matrices
2M
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}
printf("\n Enter second matrix");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
scanf("%d",&b[i][j]);
}
}
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
c[i][j]=a[i][j]+b[i][j];
}
}
printf("\n Addition:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
printf("%d\t",c[i][j]);
}
printf("\n");
}
getch();
}
Calculat
ing
addition
2M
Display
addition
1M
(c)
Ans.
Write a program to compute the sum of all elements stored in an
array using pointers.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
void main()
{
int a[5],sum=0,i,*ptr;
clrscr();
printf("\n Enter array elements:");
for(i=0;i<5;i++)
scanf("%d",&a[i]);
6M
Variable
declarati
on 1M
Input
array
1M
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ptr=&a[0];
for(i=0;i<5;i++)
{
sum=sum+(*ptr);
ptr=ptr+1;
}
printf("\n Sum= %d",sum);
getch();
}
Pointer
Initializ
ation
1M
Sum
calculati
on 2M
Display
1M
6.
(a)
Ans.
Attempt any TWO of the following:
Write a program to sort elements of an array in ascending order.
(Note: Any other correct logic shall be considered).
#include<stdio.h>
#include<conio.h>
void main()
{
int a[5],i,j,temp;
clrscr();
printf("\n Enter array elements:");
for(i=0;i<5;i++)
scanf("%d",&a[i]);
for(i=0;i<5;i++)
{
for(j=0;j<4-i;j++)
{
if(a[j]>a[j+1])
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
}
for(i=0;i<5;i++)
printf("\n %d",a[i]);
getch();
}
12
6M
Input
array
1M
Sorting
logic 4M
Display
sorted
list 1M
(b)
Write a function to print Fibonacci series starting from 0, 1.
(Note: Any other correct logic shall be considered).
6M
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Ans.
void Fibbo()
{
inta,b,c,limit,i;
printf("\n Enter number:");
scanf("%d",&limit);
a=0;
b=1;
printf("%d\t%d",a,b);
for(i=0;i<limit-2;i++)
{
c=a+b;
printf("\t%d",c);
a=b;
b=c;
}
}
Correct
function
with
syntax
6M
(c)
Ans.
Calculate factorial of a number using recursion.
(Note: Explanation/algorithm/program shall be considered)
#include<stdio.h>
#include<conio.h>
int factorial(int no)
{
if(no==1)
return(1);
else
return(no*factorial(no-1));
}
void main()
{
intfact,no;
clrscr();
printf("\n Enter number");
scanf("%d",&no);
fact=factorial(no);
printf("\n Factorial number=%d",fact);
getch();
}
6M
Recursiv
e
function
4M
Main
function
2M
