GATE CS 2016 Set 2 Q.22 (CS Q.12): Functions, Recursion, Scope and Storage Classes

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);
}

Your answer

Official answer and solution

Solution

Trace the pointer dereferencing and pass-by-value argument modifications during the function call.

In main():
int i = 5, j = 10;
f(&i, j);

Examine the function signature and execution of f(int* p, int m):

p receives the address of i (p = &i). Any modification to *p directly modifies i in main().
m receives a copy of the value of j (m = 10). This is call-by-value, so changes to m have no effect on j in main().

Inside f:

m = m + 5; $\implies m = 10 + 5 = 15$ .
*p = *p + m; $\implies *p = 5 + 15 = 20$ . Since p points to i, this sets i = 20.
return;

Back in main():

Final answer

Common trap

Assuming j was also modified by the function call (mistaking pass-by-value for pass-by-reference).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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