GATE CS 2015 Set 1 Q.21 (CS Q.11): Pointers, Arrays and Strings

The output of the following C program is _____.

#include <stdio.h>

void f1(int a, int b) {
    int c;
    c = a; a = b; b = c;
}

void f2(int *a, int *b) {
    int c;
    c = *a; *a = *b; *b = c;
}

int main() {
    int a = 4, b = 5, c = 6;
    f1(a, b);
    f2(&b, &c);
    printf("%d", c - a - b);
    return 0;
}

Your answer

Official answer and solution

Official answer: -5
From: Official GATE 2015 answer key

Solution

Trace the C function calls, distinguishing between call-by-value and call-by-reference using pointers.

In main(), variables are initialized: a = 4, b = 5, c = 6.
Call f1(a, b):
f1(int a, int b) receives arguments by value. The swaps within f1 affect only its local copies. On return, variables in main() remain unchanged: a = 4, b = 5, c = 6.
Call f2(&b, &c):
f2(int *a, int *b) receives the addresses of b and c from main():
```
int c; // local variable in f2
c = *a; // c = 5
*a = *b; // *a (main's b) = 6
*b = c;  // *b (main's c) = 5
```
Thus, after f2(&b, &c), b becomes $6$ $6$ and c becomes $5$ $5$ .
Evaluate the printf expression c - a - b: $\text{Value} = c - a - b = 5 - 4 - 6 = -5.$

The output printed is -5.

Final answer

-5

Common trap

Assuming f1 modifies main's a and b, or miscalculating the arithmetic expression $5 - 4 - 6$ as $+5$ .

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

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 21 of 65, printed as CS Q.11
Section: Computer Science and Information Technology

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