Consider the following recursive C function that takes two arguments

Consider the following recursive C function that takes two arguments

unsigned int foo(unsigned int n, unsigned int r) {   if (n  > 0) return (n%r +  foo (n/r, r ));   else return 0; }

What is the return value of the function foo when it is called as foo(513, 2)?
(A) 9
(B) 8
(C) 5
(D) 2

Answer: (D)
foo(513, 2) will return 1 + foo(256, 2). All subsequent recursive calls (including foo(256, 2)) will return 0 + foo(n/2, 2) except the last call foo(1, 2) . The last call foo(1, 2) returns 1. So, the value returned by foo(513, 2) is 1 + 0 + 0…. + 0 + 1.
The function foo(n, 2) basically returns sum of bits (or count of set bits) in the number n.