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What will be the output of the program (in Turbo C)? #include int fun(int *f) { *f = 10; return 0; } int main() { const int arr = {1, 2, 3, 4, 5}; printf("Before modification arr = %d", arr); fun(&arr); printf("\nAfter

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1. 

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

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Computer Science & Engineering :: C Programming Questions and Answers : More Exercises

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