为什么 Math.round(0.49999999999999994) 返回 1？

小编典典

为什么 Math.round(0.49999999999999994) 返回 1？

all

在下面的程序中，您可以看到每个略小于的值.5都被四舍五入，除了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);
}

印刷

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

我正在使用 Java 6 更新 31。

阅读 73

2022-03-07

共1个答案

小编典典

概括

在 Java 6（可能更早）中，round(x)实现为floor(x+0.5). 1 这是一个规范错误，正是针对这种病态的情况。2 Java 7
不再强制执行这种损坏的实现。3

问题

0.5+0.49999999999999994 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 0.49999999999999994 的指数小于 0.5，所以当它们相加时，尾数会移动，ULP 会变大。

解决方案

从 Java 7 开始，OpenJDK（例如）是这样实现的：4

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;
}

2022-03-07