为什么Math.round（0.49999999999999994）返回1？

问题描述

在下面的程序中，你可以看到每个小于 .5 的值被舍去，除了 0.5 。

pre $ for（int i = 10; i> = 0; i--）{
long l = Double.doubleToLongBits（i + 0.5）;
double x;
do {
x = Double.longBitsToDouble（l）;
System.out.println（x +rounded is+ Math.round（x））;
l--;
} while（Math.round（x）> i）;






$ b

  10.5圆整为11 
 10.499999999999998圆整为10 
圆整为10 
 9.499999999999998圆整为9 
 8.5圆整为9 
 8.499999999999998圆整是8 
 7.5圆整是8 
 7.499999999999999圆整是7 
 6.5圆整是7 
 6.499999999999999圆整是6 
 5.5圆整是6 
 5.499999999999999四舍五入5 
 4.5圆整为5 
 4.499999999999999圆整为4 
 3.5圆整为4 
 3.4999999999999996圆整为3 
 2.5圆整为3 
 2.4999999999999996圆整为2 
 1.5圆整为2 
 1.4999999999999998圆整为1 
 0.5圆整为1 
 0.49999999999999994圆整为1 
 0.4999999999999999圆整为0 
  
 
 
 
我正在使用Java 6更新版本31.   
 
解决方案
  round（x） 被实现为 floor （x + 0.5）。这是一个规范错误，恰恰是这一个病态。 2 Java 7不再强制执行这个破坏的实现 
  
 
 
 问题 
 
 0.5+ 0.49999999999999994正好是1的双精度：
 
 pre $静态无效打印（double d）{
 System.out.printf（ ％016x\\\
，Double.doubleToLongBits（d））; 
 
 
 public static void main（String args []）{
 double a = 0.5; 
 double b = 0.49999999999999994; 
 
 print（a）; // 3fe0000000000000 
 print（b）; // 3fdfffffffffffff 
 print（a + b）; // 3ff0000000000000 
 print（1.0）; // 3ff0000000000000 
} 
  

这是因为0.49999999999999994的指数小于0.5，他们被加了，它的尾数被转移了，ULP也变大了。
$ b

解决方案

从Java 7开始，OpenJDK（例如）就这样实现它：
$ b

  （double a）{
 if（a！= 0x1.fffffffffffffp-2）//最大double值小于0.5 
 return（long）floor（a + 0.5d）; 
 else 
 return 0; 
 
 $ / code> 





---

$ sub
> http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29

4.

In the following program you can see that each value slightly less than .5 is rounded down, except for 0.5.

for (int i = 10; i >= 0; i--) {
    long l = Double.doubleToLongBits(i + 0.5);
    double x;
    do {
        x = Double.longBitsToDouble(l);
        System.out.println(x + " rounded is " + Math.round(x));
        l--;
    } while (Math.round(x) > i);
}

prints

10.5 rounded is 11
10.499999999999998 rounded is 10
9.5 rounded is 10
9.499999999999998 rounded is 9
8.5 rounded is 9
8.499999999999998 rounded is 8
7.5 rounded is 8
7.499999999999999 rounded is 7
6.5 rounded is 7
6.499999999999999 rounded is 6
5.5 rounded is 6
5.499999999999999 rounded is 5
4.5 rounded is 5
4.499999999999999 rounded is 4
3.5 rounded is 4
3.4999999999999996 rounded is 3
2.5 rounded is 3
2.4999999999999996 rounded is 2
1.5 rounded is 2
1.4999999999999998 rounded is 1
0.5 rounded is 1
0.49999999999999994 rounded is 1
0.4999999999999999 rounded is 0

I am using Java 6 update 31.

解决方案

Summary

In Java 6 (and presumably earlier), round(x) is implemented as floor(x+0.5). This is a specification bug, for precisely this one pathological case. Java 7 no longer mandates this broken implementation.

The problem

0.5+0.49999999999999994 is exactly 1 in double precision:

static void print(double d) {
    System.out.printf("%016x\n", Double.doubleToLongBits(d));
}

public static void main(String args[]) {
    double a = 0.5;
    double b = 0.49999999999999994;

    print(a);      // 3fe0000000000000
    print(b);      // 3fdfffffffffffff
    print(a+b);    // 3ff0000000000000
    print(1.0);    // 3ff0000000000000
}

This is because 0.49999999999999994 has a smaller exponent than 0.5, so when they're added, its mantissa is shifted, and the ULP gets bigger.

The solution

Since Java 7, OpenJDK (for example) implements it thus:

public static long round(double a) {
    if (a != 0x1.fffffffffffffp-2) // greatest double value less than 0.5
        return (long)floor(a + 0.5d);
    else
        return 0;
}

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