How to code a modulo (%) operator in C/C++/Obj-C that handles negative numbers

Question

One of my pet hates of C-derived languages (as a mathematician) is that

(-1) % 8 // comes out as -1, and not 7

fmodf(-1,8) // fails similarly

What's the best solution?

C++ allows the possibility of templates and operator overloading, but both of these are murky waters for me. examples gratefully received.

Answer 1

First of all I'd like to note that you cannot even rely on the fact that (-1) % 8 == -1. the only thing you can rely on is that (x / y) * y + ( x % y) == x. However whether or not the remainder is negative is implementation-defined.

Now why use templates here? An overload for ints and longs would do.

int mod (int a, int b)
{
   int ret = a % b;
   if(ret < 0)
     ret+=b;
   return ret;
}

and now you can call it like mod(-1,8) and it will appear to be 7.

Edit: I found a bug in my code. It won't work if b is negative. So I think this is better:

int mod (int a, int b)
{
   if(b < 0) //you can check for b == 0 separately and do what you want
     return -mod(-a, -b);   
   int ret = a % b;
   if(ret < 0)
     ret+=b;
   return ret;
}

Reference: C++03 paragraph 5.6 clause 4:

The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined; otherwise (a/b)*b + a%b is equal to a. If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined.