GATE | GATE CS 2019 | Question 35

• Last Updated : 05 Aug, 2021

Consider the following C program:

 void convert(int n) {  if (n < 0)    printf(“ % d”, n);  else {    convert(n / 2);    printf(“ % d”, 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);}

My Personal Notes arrow_drop_up