C Programming - Const

Exercise :: Const - Find Output of Program

6. 

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;
}

A. Error
B. H
C. Hello
D. Hel

Answer: Option C

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".


7. 

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;
}

A. Garbage value
B. Error
C. 20
D. 0

Answer: Option C

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".


8. 

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

Answer: Option A

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


9. 

What will be the output of the program?

#include<stdio.h>

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

A. 10
B. 11
C. No output
D. Error: ++needs a value

Answer: Option D

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("%d\n", 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.


10. 

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;
}

A. Error
B. -11, 34
C. 11, 34
D. None of these

Answer: Option B

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, %d\n", c, d); The value of the variable 'c' and 'd' are printed.

Hence the output of the program is -11, 34


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