Consider the same recursive C++ function that takes two arguments
C++
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)
Explanation:
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.
Hence (D) is the correct Answer.
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