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问题：

在下面的程序中，您可以看到，除了0.5之外，每个略小于.5的值都向下舍入。

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);
}

结果：

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

我使用的java版本是Java 6 update 31.

热门答案：

答案1：

总结：
【1】-在Java6（以及更早的版本）中，round（x）被实现为floor（x+0.5）。
【2】-这是一个规范错误，正是针对这一病态情况。
【3】-Java7不再强制要求这种中断的实现。
问题：
0.5+0.4999999994在双精度中正好是1：

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0.49999994的指数比0.5小，所以当它们相加时，它的尾数会移动，ULP会变大。 解决方案：
【4】-从Java7开始，OpenJDK（例如）就这样实现了:

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】.docs.oracle.com/javase/6/do…

• 【2】.bugs.java.com/bugdatabase… (credits to @SimonNickerson for finding this)

• 【3】.docs.oracle.com/javase/7/do…

• 【4】.grepcode.com/file/reposi…

文章翻译自Stack Overflow