浮点精度 [英] Floating point accuracy

问题描述

我正在写一种方法来计算 a * x + b * y = 1

  //给定两个点，通过求解两个线性方程来求出线的方程，然后测试结果。 （为简单起见，假设delta！= 0）
 
 private boolean solveAndRetry（float x1，float y1，float x2，float y2）{
 float delta = x1 * y2  -  x2 * Y1; 
 float deltaA = y2  -  y1; 
 float deltaB = x1  -  x2; 
 
 float a = deltaA / delta; 
 float b = deltaB / delta; 
 float c = 1; 
 
 // test 
 if（a * x2 + b * y2 == c）{
 System.out.println（ok）; 
返回true; 
} 
 else {
 System.out.println（a * x2 + b * y2-c）; 
返回false; 
 
 
 
 
 $ b当我运行它的时候，所有确定，但事实并非如此，我不知道为什么
  public static void main（String [] （float x = 0; x <10; x + = 0.01f）{
 solveAndRetry（1，-1，x，2）; 
 
 
 
 
 
 
 
结果如下：
 
 
 pre $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ ok 
 ok 
 -5.9604645E-8 
 ok 
 -5.9604645E-8 
 ok 
 ok 
 ok 
 1.1920929E -7 
 ok 
 ok 
 -5.9604645E-8 
 ok 
  
 
 
 float 的精确度为6到7位十进制数字。由于四舍五入的错误无法避免，所以你的结果是可以得到的。
 
 
 
通常，你永远不会比较浮点数的相等性。   $ b 
  > Math.abs（x-y） eps 
  
为适当选择的eps。
 
I am writing a method which calculates the equation of a 2D Line in the form a*x+b*y=1
//Given two points, find the equation of the line by solving two linear equations and then test the result. (For simplicity, assume that delta !=0 here)

private boolean solveAndRetry(float x1,float y1, float x2,float y2) {
        float delta = x1 * y2 - x2 * y1;
        float deltaA = y2 - y1;
        float deltaB = x1 - x2;

        float a = deltaA / delta;
        float b = deltaB / delta;
        float c = 1;

        //test
        if (a * x2 + b * y2 == c) {
        System.out.println("ok");
            return true;
        }
        else {
            System.out.println(a * x2 + b * y2-c);
            return false;
        }
    }
When I ran it, I was expecting there would be all "ok"s, but that's not the case and I don't know why
public static void main(String[] args) {
        for (float x = 0; x < 10; x += 0.01f) {
            solveAndRetry(1, -1, x, 2);
        }
    }
Here are some lines in the result
ok
ok
ok
ok
ok
ok
ok
ok
-5.9604645E-8
ok
-5.9604645E-8
ok
ok
ok
1.1920929E-7
ok
ok
-5.9604645E-8
ok

解决方案
A float has an accuracy of 6 to 7 decimal digits. Since rounding errors cannot be avoided, your results are as good as it can get.

Normally, you would never compare floating point numbers for equality. Instead of x == y always use a comparison with an interval:
Math.abs(x - y) < eps
for a suitably chosen eps.

                        
这篇关于浮点精度的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！