Const Questions and Answers updated daily – C Language



10 Const Questions and answers section with explanation for various online exam preparation, various interviews, C Language Const online test. Const Questions with detailed description, explanation will help you to master the topic.

Const Questions

1.
  What will be the output of the program?

#include

int main()
{
    const c = -11;
    const int d = 34;
    printf("%d, %dn", c, d);
    return 0;
}



SHOW ANSWER
Correct Ans:-11,34
Explanation:
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


2.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const int i=0;
    printf("%dn", i++);
    return 0;
}



SHOW ANSWER
Correct Ans:Error: ++needs a value
Explanation:
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.


3.
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;
}



SHOW ANSWER
Correct Ans:Before modification arr[3] = 4 After modification arr[3] = 10
Explanation:
Step 1: const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized to

arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5

Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4).

Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10.

A const variable can be indirectly modified by a pointer.

Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10).

Hence the output of the program is

Before modification arr[3] = 4

After modification arr[3] = 10


4.
What will be the output of the program?

#include<stdio.h>
int get();

int main()
{
    const int x = get();
    printf("%d", x);
    return 0;
}
int get()
{
    return 20;
}



SHOW ANSWER
Correct Ans:20
Explanation:
Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.

Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".

The function get() returns the value "20".

Step 3: printf("%d", x); It prints the value of the variable x.

Hence the output of the program is "20".


5.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const char *s = "";
    char str[] = "Hello";
    s = str;
    while(*s)
        printf("%c", *s++);

    return 0;
}



SHOW ANSWER
Correct Ans:hello
Explanation:
Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.

Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".

Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".

Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.

Hence the output of the program is "Hello".


6.
What will be the output of the program in TurboC?

#include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10, j=20;
    const int *ptr = &i;
    printf(" i = %5X", ptr);
    printf(" ptr = %d", *ptr);
    ptr = &j;
    printf(" j = %5X", ptr);
    printf(" ptr = %d", *ptr);
    return 0;
}



SHOW ANSWER
Correct Ans:i= FFE2 ptr=10 j=FFE4 ptr=20
Explanation:


7.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const int x=5;
    const int *ptrx;
    ptrx = &x;
    *ptrx = 10;
    printf("%dn", x);
    return 0;
}



SHOW ANSWER
Correct Ans:error
Explanation:
Step 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'.

Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer.

Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx.

Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error.

To change the value of const variable x we have to use *(int *)&x = 10;


8.
What will be the output of the program?

#include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10;
    const int *ptr = &i;
    fun(&ptr);
    return 0;
}
int fun(int **ptr)
{
    int j = 223;
    int *temp = &j;
    printf("Before changing ptr = %5xn", *ptr);
    const *ptr = temp;
    printf("After changing ptr = %5xn", *ptr);
    return 0;
}



SHOW ANSWER
Correct Ans:Error: cannot convert parameter 1 from 'const int **' to 'int **'
Explanation:


9.
What will be the output of 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 %d %f", e1.name, e1.age, e1.salary);
    return 0;
}



SHOW ANSWER
Correct Ans:No error
Explanation:
The output will be (in 16-bit platform DOS):

K 75 0.000000


10.
What will be the output of the program?

#include<stdio.h>

int main()
{
    int y=128;
    const int x=y;
    printf("%dn", x);
    return 0;
}



SHOW ANSWER
Correct Ans:128
Explanation:
Step 1: int y=128; The variable 'y' is declared as an integer type and initialized to value "128".

Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value.

Step 3: printf("%dn", x); It prints the value of variable 'x'.

Hence the output of the program is "128"


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