浮点循环运行的迭代次数应该比应有的多

Question

I have this loop thats supposed to iterate lx times over x (range [-1,1]) and ly times over y (range [-1,1]), 9 times total, yet it iterates 16 times! Why is this??

#include <stdio.h>

int main(void) {
  double lx=3,ly=3;
  double du = (1-(-1))/double(lx);
  double dv = (1-(-1))/double(ly);
  int count=0;

  for (double y=-1; y < 1; y += dv)
    for (double x=-1; x < 1; x += du) {

      count++;
    }

  printf("\n\n%d lss\n", count);
}

Answer 1

Seriously?

You are iterating over [-1, -1/3, 1/3, 1].

You can see this by putting an output statement in the loop to see what's happening.

Answer 2

I assume your C implementation use IEEE-754 64-bit binary floating-point for double , which is the most common choice.

Every numerical arithmetic system is inexact, so the numbers you work with are not the same numbers you would get with exact mathematics.

The statement double du = (1-(-1))/double(lx); sets du to the double value that is closest to 2/3, which is 0.66666666666666662965923251249478198587894439697265625.

In the first iteration of the outer loop, y is -1. When du is added to it at the end of each iteration, the results are:

• -0.33333333333333337034076748750521801412105560302734375,
• 0.3333333333333332593184650249895639717578887939453125, and
• 0.99999999999999988897769753748434595763683319091796875.

Obviously, that last value of y is less than 1, so the loop continues for another iteration.

Loop iterators should usually use integer values, so that it is easy to ensure exact arithmetic is used. You can scale the iterator to produce other desired values:

for (int iy = 0; iy < 4; ++iy)
{
    double y = iy * 2 / 3. - 1;
    …
}

Answer 3

You need to calculate the difference between your start and end point in a different way here as you're using doubles here.

So instead of end - start use end - start + 1.0 and you're done.