C++ integer floor function

5/3 will always produce 1 (an integer), if you do 5.0/3 or 5/3.0 the result will be a double.

As far as I know, there is no predefined function for this purpose. It might be necessary to use such a function, if for some reason floating-point calculations are out of question (e.g. int64_t has a higher precision than double can represent without error)

We could define this function as follows:

#include <cmath>

inline long
floordiv (long num, long den)
{
  if (0 < (num^den))
    return num/den;
  else
    {
      ldiv_t res = ldiv(num,den);
      return (res.rem)? res.quot-1 
                      : res.quot;
    }
}

The idea is to use the normal integer divison, but adjust for negative results to match the behaviour of the double floor(double) function. The point is to truncate always towards the next lower integer, irrespective of the position of the zero point. This can be very important if the intention is to create even sized intervals.

Timing measurements show that this function here only creates a small overhead compared with the built-in / operator, but of course the floating point based floor function is significantly faster....

You will lose the fractional portion of the quotient. So yes, with greater numbers you will have more relative precision, such as compared with 5000/3000.

However, 5 / 3 will return an integer, not a double. To force it to divide as double, typecast the dividend as static_cast<double>(5) / 3.

Integer division gives integer results, so 5 / 3 is 1 and 5 % 3 is 2 (the remainder operator). However, this doesn't necessarily hold with negative numbers. In the original C++ standard, -5 / 3 could be either -1 (rounding towards zero) or -2 (the floor), but -1 was recommended. In the latest C++0B draft (which is almost certainly very close to the final standard), it is -1, so finding the floor with negative numbers is more involved.