# GATE | GATE CS 2019 | Question 35

• Last Updated : 05 Aug, 2021

Consider the following C program:

 `void` `convert(``int` `n) {``  ``if` `(n < 0)``    ``printf``(ā % dā, n);``  ``else` `{``    ``convert(n / 2);``    ``printf``(ā % dā, n % 2);``  ``}``}`

Which one of the following will happen when the function convert is called with any positive integer n as argument?
(A) It will print the binary representation of n in the reverse order and terminate.
(B) It will print the binary representation of n but will not terminate
(C) It will not print anything and will not terminate.
(D) It will print the binary representation of n and terminate.

Explanation: Since n is the integer, so it 1/2 = 0.5 = 0 will return because of integer.

0/2 = 0 will cause infinite loop because there is no terminating condition for 0.

So, option (C) is correct.

Note:
It will print the binary representation of n and terminate, only if condition “if (n <= 0)".

 `#include `` ` `void` `convert(``int` `n) {`` ` `  ``if``(n <= 0)``  ``printf``(``"%d"``, n);``   ` `  ``else` `{``  ``convert(n / 2);``  ``printf``(``"%d"``, n%2);``  ``};``}`` ` `int` `main() {`` ``convert (16);``}`

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