Question

what did I do wrong in this functions. I am pretty sure the problem is at base = exp(base, pwr /= 2) * exp(base, pwr /= 2); but I cannot think of a logical reason. is there a possible way to write a parameter like that? thanks in advance. (p.s my output of this function is a 2 which is wrong)

#include <iostream>
using namespace std;

unsigned long& exp(unsigned long& base, unsigned long& pwr)
{
    if(pwr == 0)
      base = 1;
    else if(pwr == 1)
      base = base;
    else
      base = exp(base, pwr /= 2) * exp(base, pwr /= 2);
    return base;
}

int main()
{
    unsigned long n=2, m = 4;
    cout << exp(n,m) << endl;
    return 0;
}    

Answer 1

Here are five things to note about this line:

  base = exp(base, pwr /= 2) * exp(base, pwr /= 2);

1. As noted in a comment above, base is passed by reference, not value, so there is only one copy of it and you’re changing its value. This is a bad idea.
2. pwr is also passed by reference and you’re changing its value when you use /= instead of just /. There are two /= statements in this line, so after this line runs, pwr now has one fourth its original value.
3. The exp function will get run twice each time you run this line. It would make more sense to store the value and then square it.
4. /2 is integer division, so it will round down. So if you give it a number like 3 as an exponent, it will not work correctly because 3/2 is 1. If you correct the other mistakes and the call it with an exponent of 7, it will end up only doing exp(2,7) = exp(2,3)*exp(2,3) = exp(2,1)*exp(2,1)*exp(2,1)*exp(2,1) = 16 when obviously the correct answer is 128. This function, as designed, will only work correctly when the exponent is a power of 2.
5. Good thing that number 4 is true because if you did get exp(2,1.5) you’d never terminate since it wouldn’t match your base cases. You should probably rethink your algorithm generally.