Official PYQ · 2017Set 1 · Q36RecursionMCQ1 mark

GATE CS 2017 Set 1 Q36 — Recursion

Consider the C functions foo and bar given below:

int foo(int val) {
    int x=0;
    while(val > 0) {
        x = x + foo(val--);
    }
    return val;
}

int bar(int val) {
    int x = 0;
    while(val > 0) {
        x= x + bar(val-1);
    }
    return val;
}

Invocations of $foo(3)$ and $bar(3)$ will result in:

View worked solution

Answer: C

Let us analyze the execution of both functions starting with the input 3:

1. Execution of foo(3):

Inside foo(3), the loop condition val > 0 (3 > 0) is true.
The statement x = x + foo(val--) is executed.
The post-decrement operator val-- passes the current value of val (which is 3) to the recursive call first, and then decrements val to 2.
This results in a recursive call to foo(3).
This process repeats indefinitely, causing infinite recursion. Since each recursive call consumes stack space, it quickly leads to a stack overflow, resulting in abnormal termination.

2. Execution of bar(3):

Inside bar(3), val = 3. It calls bar(2), which calls bar(1), which in turn calls bar(0).
bar(0) returns 0 immediately because the loop condition val > 0 is false.
Control returns to bar(1). The statement x = x + bar(0) completes, and x becomes 0.
The loop condition val > 0 is checked again in bar(1). Since val is still 1 (it was never decremented or modified inside bar), the condition remains true.
This causes bar(1) to call bar(0) repeatedly in an infinite loop without growing the stack indefinitely.

Thus, foo(3) terminates abnormally, and bar(3) enters an infinite loop.

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Published 10 Oct 2026

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