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

Question

在以下程序中,您可以看到每个值略小于.5向下舍入,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);
}

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更新31.

Answer 1

摘要

在Java 6中(可能更早),round(x)实现为floor(x+0.5).¹ 这是一个规范错误,正是这一个病态案例.² Java 7不再要求这种破坏的实现.³

问题

0.5 + 0.49999999999999994正是双精度中的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,因此当它们被添加时,它的尾数被移位,并且ULP变大.

解决方案

从Java 7开始,OpenJDK(例如)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;
}

_{1. 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(致@SimonNickerson发现此信息)}

_{3. 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}

Answer 2

这似乎是一个已知的错误(Java bug 6430675:Math.round对0x1.fffffffffffffp-2有令人惊讶的行为)已在Java 7中得到修复.

Answer 3

JDK 6中的源代码:

public static long round(double a) {
    return (long)Math.floor(a + 0.5d);
}

JDK 7中的源代码:

public static long round(double a) {
    if (a != 0x1.fffffffffffffp-2) {
        // a is not the greatest double value less than 0.5
        return (long)Math.floor(a + 0.5d);
    } else {
        return 0;
    }
}

当值为0.49999999999999994d时,在JDK 6中,它将调用floor并因此返回1,但在JDK 7中,if条件是检查该数字是否是小于0.5的最大double值.在这种情况下,数字不是小于0.5的最大双精度值,因此该else块返回0.

您可以尝试0.49999999999999999d,它将返回1,但不返回0,因为这是小于0.5的最大双精度值.

Answer 4

我在JDK 1.6 32位上也一样,但在Java 7 64位上,我得到0为0.49999999999999994,舍入为0,最后一行不打印.这似乎是一个VM问题,但是,使用浮点,您应该期望在各种环境(CPU,32位或64位模式)上的结果有所不同.

并且,当使用round或反转矩阵等时,这些位可以产生巨大的差异.

x64输出:

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 0

Answer 5

以下的答案是Oracle 错误报告6430675的摘录.请访问该报告以获取完整说明.

方法{Math,StrictMath.round在操作上定义为

(long)Math.floor(a + 0.5d)

用于双重参数.虽然这个定义通常按预期工作,但它给出了令人惊讶的结果,而不是0,对于0x1.fffffffffffffp-2(0.49999999999999994).

值0.49999999999999994是小于0.5的最大浮点值.作为十六进制浮点字面值,其值为0x1.fffffffffffffp-2,等于(2 - 2 ^ 52)*2 ^ -2.==(0.5 - 2 ^ 54).因此,总和的确切值

(0.5 - 2^54) + 0.5

是1 - 2 ^ 54.这是两个相邻浮点数(1 - 2 ^ 53)和1之间的中间点.在IEEE 754算术轮到Java使用的最接近舍入模式时,当浮点结果不精确时,两者的距离越近必须返回包含确切结果的可表示浮点值; 如果两个值都相等,则返回其最后一位为零的值.在这种情况下,add的正确返回值是1,而不是小于1的最大值.

虽然该方法按照定义运行,但此输入的行为非常令人惊讶; 规范可以修改为更像"Round to the long long,rounding tied up",这将允许更改此输入的行为.