GATE CS 2012 Q.48 (CS Q.48): Functions, Recursion, Scope and Storage Classes

Common data questions

Common Data for Questions 48 and 49:

Consider the following C code segment.

int a, b, c = 0;
void prtFun(void);
main() {
    static int a = 1; /* Line 1 */
    prtFun();
    a += 1;
    prtFun();
    printf("\n %d %d ", a, b);
}

void prtFun(void) {
    static int a = 2; /* Line 2 */
    int b = 1;
    a += ++b;
    printf("\n %d %d ", a, b);
}

Shared with question 49.

What output will be generated by the given code segment?

3 1
4 1
4 2

4 2
6 1
6 1

4 2
6 2
2 0

3 1
5 2
5 2

Official answer and solution

Official answer: C
From: Official GATE 2012 answer key

Solution

Trace variable scopes and static variable retention across function calls in C.

Initial declarations:

Global: int a = 0, b = 0, c = 0;
main() defines local static int a = 1;.
prtFun() defines local static int a = 2; and auto int b = 1;.

Execution trace:

Call 1 to prtFun():
- Inside prtFun: static local a = 2, local b = 1.
- ++b increments local b to 2.
- a += ++b sets a = 2 + 2 = 4.
- Prints 4 2.
Back in main():
- a += 1: main's static a becomes $1 + 1 = 2$ .
Call 2 to prtFun():
- Inside prtFun: static a retains 4 from previous call. Auto b is reinitialized to 1.
- ++b increments local b to 2.
- a += ++b sets a = 4 + 2 = 6.
- Prints 6 2.
Back in main():
- printf("\n %d %d ", a, b);
- Local static a in main is 2.
- b is not declared locally in main, so it accesses global b, which is 0.
- Prints 2 0.

Output sequence:
4 2
6 2
2 0
This matches option (C).

Final answer

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

Source

Paper: GATE 2012 Computer Science and Information Technology (CS)
Question: 48 of 65, printed as CS Q.48
Section: Computer Science and Information Technology

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