Question

The value printed by the following program is _______.

void f (int* p, int m) {

m = m + 5;

*p = *p + m;

return;

}

void main () {

        int i=5, j=10;

        f (&i, j);

        printf (‘’%d’’, i+j);

}

Answer 1

Code Explanation:

In this code, address of i is passed to the function and value of j is passed.

Given i =5, j= 10

void f (int* p, int m) { //here p points to address of i and m contains 5

            m = m + 5; // m becomes 15

            *p = *p + m; // p becomes 20

            return;

}

When call returns back to the main function, i value changes to 20 but j will remain same.

So, printf (‘’%d’’, i + j); // 20 + 10 = 30 is printed