Question
#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
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 D
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Answer: Option A
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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