Merge pull request giantray#79 from 0p3rator/master

why-does-math-round0-49999999999999994-return-1问题翻译

Author: giantray

contents/why-does-math-round0-49999999999999994-return-1.md

@@ -0,0 +1,96 @@
+# 为什么数学函数Math.round(0.49999999999999994) 返回 1
+
+tags:stackoverflow-java-top-qa
+
+---
+ 
+###问题
+通过下面的程序你可以看出来,对于任意一个比0.5略小的都是舍去小数向下取整,只有0.5是例外.
+
+```java
+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);
+}
+```
+
+输出为:
+
+```
+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
+
+```
+*_译者注:请看输出的最后两行,0.49999999999999994的输出为1,而0.49999999999999999的输出为0*
+
+我使用的版本是 Java 6 update 31
+
+### 回答
+**总结**
+
+在 Java 6(或者之前的版本),round(x)是用floor(x+0.5)实现的.¹ 这是一个规范上的bug,恰恰是在这种病理条件下.²Java 7 不再使用这个有问题的实现了.
+
+**问题**
+
+0.5+0.49999999999999994 在double的精度下的结果是1
+```java
+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
+}
+```
+这是因为0.49999999999999994的指数比0.5的指数小,所以当它们两个相加时,0.49999999999999994的原数就会发生移位,然后最小精度单位(unit of least precision)/最后置单位(unit of last place)相应的变大了.
+
+**解决方案**
+
+自从Java 7以来,OpenJDK(举个栗子)实现如下⁴:
+
+```java
+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;
+}
+```
+1. [http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#round%28double%29](http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#round%28double%29)
+2. [http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675](http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675)  (credits to @SimonNickerson for finding this)
+3. [http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29](http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29)
+4. [http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7u40-b43/java/lang/Math.java#Math.round%28double%29](http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7u40-b43/java/lang/Math.java#Math.round%28double%29)
+
+### stackoverflow原文链接:[http://stackoverflow.com/questions/9902968/why-does-math-round0-49999999999999994-return-1](http://stackoverflow.com/questions/9902968/why-does-math-round0-49999999999999994-return-1)