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 : A) Return of 6 and 6 respectively. B) Infinite loop and abnormal termination respectively. C) Abnormal termination and infinite loop respectively. D) Both terminating abnormally.

Official answer: C) Abnormal termination and infinite loop respectively.. Explanation: Let us analyze the behavior of both functions: 1. For `foo(3)`: - When `foo(3)` is called, `val = 3 > 0` is true. - Inside the loop, it evaluates `foo(val--)`. The postfix decrement `val--` evaluates to 3 as the argument to the recursive call, before decrementing `val`. - Therefore, `foo(3)` immediately calls `foo(3)` again. - This causes an infinite sequence of recursive calls with the parameter 3, consuming stack frames continuously until memory runs out. - This results in a stack overflow, leading to abnormal termination (crash). 2. For `bar(3)`: - When `bar(3)` is called, it enters `while(val > 0)` and calls `bar(2)`, which in turn calls `bar(1)`. - Inside `bar(1)`: - The loop condition `val > 0` (1 > 0) is true. - It calls `bar(0)`. - Inside `bar(0)`, `val = 0`, so the loop `while(val > 0)` is skipped and `bar(0)` immediately returns 0. - Control returns to `bar(1)`, where `x = x + 0` is executed. - In `bar(1)`, the value of `val` is still 1 because `val - 1` was passed by value and `val` itself was never modified. - The loop in `bar(1)` continues with `val = 1 > 0` and calls `bar(0)` repeatedly in an infinite loop without causing a stack overflow (as `bar(0)` returns immediately each time). Hence, `foo(3)` results in abnormal termination, and `bar(3)` results in an infinite loop. Thus, option (C) is the correct answer.

A) Return of 6 and 6 respectively.

B) Infinite loop and abnormal termination respectively.

C) Abnormal termination and infinite loop respectively.

D) Both terminating abnormally.

Programming and Data Structures · Recursion

Official IIT answer key · IIT Roorkee · Audited Aug 2026

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GATE CSE 2017 Set 1 Q36 · MCQ · 2 marks

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 :

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Recursion

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How GATE tests Recursion

GATE tests Recursion primarily through direct execution tracing of recursive C code snippets, stack-unwinding behavior (pre-order vs. post-order execution), and termination/infinite recursion analysis. Questions frequently incorporate static variables, pointer/reference aliasing, mutual recursion, and tree/array traversals to test whether candidates can systematically maintain call frames. In recent years, the focus has expanded from standard tracing to higher-order mathematical characterizations (e.g., finding the smallest input or maximum output value of composed recurrences).

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Programming and Data Structures · Recursion

Official IIT answer key · IIT Roorkee · Audited Aug 2026

Take topic test

GATE CSE 2017 Set 1 Q36 · MCQ · 2 marks

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 :

Select an option to check

Key concept

Recursion

Step-by-step verified solution

Examiner trap & misconception

Topic notes

How GATE tests Recursion

Full Recursion guide →

More on Recursion

All 19 questions on Recursion →

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

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 :

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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 :

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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 :

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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 :

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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 :

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 :