Questions and answers for exam Preparation

MCQs

Total Questions : 16 | Page 1 of 2 pages

Question 1.

Point out the error in the program.

#include<stdio.h>
const char *fun();
int main()
{
*fun() = 'A';
return 0;
}
const char *fun()
{
return "Hello";
}

Error: RValue required
Error: Lvalue required
Error: fun() returns a pointer const character which cannot be modified
No error

Discuss Question

Answer: Option C. -> Error: fun() returns a pointer const character which cannot be modified

Question 2.

Point out the error in the program.

#include<stdio.h>
int main()
{
const int k=7;
int *const q=&k;
printf("%d", *q);
return 0;
}

Error: RValue required
Error: Lvalue required
Error: cannot CONVERT from 'const int *' to 'int *const'
No error

Discuss Question

Answer: Option D. -> No error

No error. This will produce 7 as output.

Question 3.

Point out the error in the program.

#include<stdio.h>
int main()
{
const int x;
x=128;
printf("%d\n", x);
return 0;
}

Error: unknown data type const int
Error: const variable have been initialised when declared.
Error: stack overflow in x
No error

Discuss Question

Answer: Option B. -> Error: const variable have been initialised when declared.

A const variable has to be initialized when it is declared. later assigning the value to the const

variable will result in an error "Cannot modify the const object".

Hence Option B is correct

Question 4.

Point out the error in the program.

#include<stdio.h>
const char *fun();
int main()
{
char *ptr = fun();
return 0;
}
const char *fun()
{
return "Hello";
}

Error: Lvalue required
Error: cannot convert 'const char *' to 'char *'
No error and No output
None of above

Discuss Question

Answer: Option C. -> No error and No output

Question 5.

Point out the error in the program.

#include<stdio.h>
#include<stdlib.h>
union employee
{
char name[15];
int age;
float salary;
};
const union employee e1;
int main()
{
strcpy(e1.name, "K");
printf("%s", e1.name);
e1.age=85;
printf("%d", e1.age);
printf("%f", e1.salary);
return 0;
}

Error: RValue required
Error: cannot modify const object
Error: LValue required in strcpy
No error

Discuss Question

Answer: Option B. -> Error: cannot modify const object

Question 6.

Point out the error in the program (in Turbo-C).

#include<stdio.h>
#define MAX 128
int main()
{
const int max=128;
char array[max];
char string[MAX];
array[0] = string[0] = 'A';
printf("%c %c\n", array[0], string[0]);
return 0;
}

Error: unknown max in declaration/Constant expression required
Error: invalid array string
None of above
No error. It prints A A

Discuss Question

Answer: Option A. -> Error: unknown max in declaration/Constant expression required

Step 1: A macro named MAX is defined with value 128

Step 2: const int max=128; The constant variable max is declared as an integer data type and it is

initialized with value 128.

Step 3: char array[max]; This statement reports an error "constant expression required". Because,

we cannot use variable to define the size of array.

To avoid this error, we have to declare the size of an array as static. Eg. char array[10];or use macro

char array[MAX];

Note: The above program will print A A as output in Unix platform.

Question 7.

What will be the output of the program?

#include<stdio.h>
int main()
{
const c = -11;
const int d = 34;
printf("%d, %d\n", c, d);
return 0;
}

Error
-11, 34
11, 34
None of these

Discuss Question

Answer: Option B. -> -11, 34

Step 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11".

Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.

Step 3: printf("%d, %dn", c, d); The value of the variable 'c' and 'd' are printed.

Hence the output of the program is -11, 34

Question 8.

What will be the output of the program?

#include<stdio.h>
int main()
{
const int i=0;
printf("%d\n", i++);
return 0;
}

10
11
No output
Error: ++needs a value

Discuss Question

Answer: Option D. -> Error: ++needs a value

This program will show an error "Cannot modify a const object".

Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value

of '0'(zero).

Step 2: printf("%dn", i++); Here the variable 'i' is increemented by 1(one). This will create an error

"Cannot modify a const object".

Because, we cannot modify a const variable.

Question 9.

What will be the output of the program (in Turbo C)?

#include<stdio.h>
int fun(int *f)
{
*f = 10;
return 0;
}
int main()
{
const int arr[5] = {1, 2, 3, 4, 5};
printf("Before modification arr[3] = %d", arr[3]);
fun(&arr[3]);
printf("\nAfter modification arr[3] = %d", arr[3]);
return 0;
}
A. Before modification arr[3] = 4
After modification arr[3] = 10
B. Error: cannot convert parameter 1 from const int * to int *
C. Error: Invalid parameter
D. Before modification arr[3] = 4
After modification arr[3] = 4