Java 为什么Math.round（0.49999999999999994）返回1？

一尘不染

Java 为什么Math.round（0.49999999999999994）返回1？

java

在下面的程序中，你可以看到.5除以外的每个值都略小于四舍五入的值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);
}

版画

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 update 31。

阅读 375

2020-03-17

共1个答案

一尘不染

摘要

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，恰恰是这种病理情况。2 Java 7不再强制执行此无效的实现。3

问题

0.5 + 0.499999999999999999994的双精度正好为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，因此当添加它们时，尾数会移动，并且ULP会变大。

解决方案

从Java 7开始，例如，OpenJDK实现了它：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;
}

2020-03-17