GATE CS 2017 Set 1 — Question 35

MCQ+2 / -0.67MediumFunctions & RecursionC ProgrammingProgramming & Data Structures

Programming & Data Structures → C Programming → Functions & Recursion

Last updated 5 September 2026

Question

Consider the following two functions. void fun1(int n) { if(n == 0) return; printf("%d", n); fun2(n - 2); printf("%d", n); } void fun2(int n) { if(n == 0) return; printf("%d", n); fun1(++n); printf("%d", n); } The output printed when fun1(5) is called is

Correct answer (A) 53423122233445 Solution Let's trace the execution of fun1(5) : 1. fun1(5) called. Prints 5 . Calls fun2(5 - 2) -> fun2(3) . 2. fun2(3) called. Prints 3 . Calls fun1(++n) -> fun1(4) (Note: n becomes 4 in fun2's scope before passing). 3. fun1(4) called. Prints 4 . Calls fun2(4 - 2) -> fun2(2) . 4. fun2(2) called. Prints 2 . Calls fun1(++n) -> fun1(3) (Note: n becomes 3). 5. fun1(3) called. Prints 3 . Calls fun2(3 - 2) -> fun2(1) . 6. fun2(1) called. Prints 1 . Calls fun1(++n) -> fun1(2) (Note: n becomes 2). 7. fun1(2) called. Prints 2 . Calls fun2(2 - 2) -> fun2(0) . fun2(0) returns immediately. Prints 2 (after return). Returns. 8. Back in fun2(1) : Prints 1 . Returns. 9. Back in fun1(3) : Prints 3 . Returns. 10. Back in fun2(2) : Prints 2 . Returns. 11. Back in fun1(4) : Prints 4 . Returns. 12. Back in fun2(3) : Prints 3 . Returns. 13. Back in fun1(5) : Prints 5 . Returns. Concatenating the printed values: 5 3 4 2 3 1 2 2 1 3 2 4 3 5 . Result: 53423122132435. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (A) 53423122233445 Solution Let's trace the execution of fun1(5) : 1. fun1(5) called. Prints 5 . Calls fun2(5 - 2) -> fun2(3) . 2. fun2(3) called. Prints 3 . Calls fun1(++n) -> fun1(4) (Note: n becomes 4 in fun2's scope before passing). 3. fun1(4) called. Prints 4 . Calls fun2(4 - 2) -> fun2(2) . 4. fun2(2) called. Prints 2 . Calls fun1(++n) -> fun1(3) (Note: n becomes 3). 5. fun1(3) called. Prints 3 . Calls fun2(3 - 2) -> fun2(1) . 6. fun2(1) called. Prints 1 . Calls fun1(++n) -> fun1(2) (Note: n becomes 2). 7. fun1(2) called. Prints 2 . Calls fun2(2 - 2) -> fun2(0) . fun2(0) returns immediately. Prints 2 (after return). Returns. 8. Back in fun2(1) : Prints 1 . Returns. 9. Back in fun1(3) : Prints 3 . Returns. 10. Back in fun2(2) : Prints 2 . Returns. 11. Back in fun1(4) : Prints 4 . Returns. 12. Back in fun2(3) : Prints 3 . Returns. 13. Back in fun1(5) : Prints 5 . Returns. Concatenating the printed values: 5 3 4 2 3 1 2 2 1 3 2 4 3 5 . Result: 53423122132435.

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