Exact double division

Question

Consider the following function:

auto f(double a, double b) -> int
{
  return std::floor(a/b);
}

So I want to compute the largest integer k such that k * b <= a in a mathematical sense. As there could be rounding errors, I am unsure whether the above function really computes this k . I do not worry about the case that k could be out of range. What is the proper way to determine this k for sure?

Answer 1

Division produces a/b, calculated exactly then rounded to a valid double number. If a/b >= n then it cannot be rounded to something less than n. If a/b < n, say 7.9999999999, then it could be rounded to 8 a priori, but you can prove that n is never the nearest floating point number.

Answer 2

The problem is that float division is not exact.

a/b can give 1.9999 instead of 2 , and std::floor can then give 1 .

One simple solution is to add a small value prior calling std::floor :

std::floor (a/b + 1.0e-10);

Result:

result = 10 while 11 was expected
With eps added, result = 11

Test code:

#include <iostream>
#include <cmath>

int main () {
    double b = atan (1.0);
    int x = 11;
    double a = x * b;
    int y = std::floor (a/b);
    std::cout << "result = " << y << " while " << x << " was expected\n";
    
    double eps = 1.0e-10;
    int z = std::floor (a/b + eps);
    std::cout << "With eps added, result = " << z << "\n";
    
    return 0;
}