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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; }
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- By: guest on 01 Jun 2017 06.00 pmStep 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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