Consider the following C function.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?

Let us trace the function convert(n) when called with a positive integer $n > 0$:

1. In the initial call with $n > 0$, the condition n < 0 is false, so it enters the else branch.

2. In the else branch, it first executes convert(n/2) before printing anything.

3. Repeated integer division by 2 eventually reduces the argument to 0 (n = 0).

4. When convert(0) is called:

   • 0 < 0 is false.

   • The else branch is entered, which immediately calls convert(0/2), i.e., convert(0) again.

5. This leads to infinite recursive calls of convert(0) because the base case condition is n < 0 instead of n <= 0 or n == 0.

6. Since the recursive call convert(n/2) appears before printf("%d", n%2), no printf statement is ever reached before the program exhausts its stack space (stack overflow).

Therefore, the function will not print anything and will not terminate, making (D) the correct option.