return value meaning | pattern

Question

What does the following function return? (in terms of meaning)

int f(int n){
    if(n == 0) return 0;
    else return n % 2 + f(n / 2)
}

Tried running the code, but couldn't find any pattern in the results

Answer 1

This output of this function can be interpreted as the number of 1s (in base 10) when the number n is represented in binary.

The base case is when n == 0 , where the number of 1s is 0 .

For every other n , there is a recursive call. There are two parts to this. The first, which is n % 2 , finds out the last bit of n . If it is a 1 , it contributes to the value returned, and is hence, counted. The second part, f(n/2) , is computing the number of 1s in all the bits of n except the last bit. This is because n/2 is n with a one bit right shift.

To put it together, the function works as follows. It checks the last bit, and if 1 , it adds it to the total. It then performs a recursive call on itself with the last bit removed. This goes on till all bits are removed, which is covered by the base case.