問題描述

在下面的程序中你可以看到每個略小于.5的值都被四舍五入了，除了0.5.

In the following program you can see that each value slightly less than .5 is rounded down, except for 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);
}

打印

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.

I am using Java 6 update 31.

推薦答案

總結(jié)

在 Java 6(可能更早)中，round(x) 被實現(xiàn)為 floor(x+0.5).¹ 這是一個規(guī)范錯誤，正是針對這種病態(tài)的情況.²Java 7 不再強制執(zhí)行這種損壞的實現(xiàn).³

In Java 6 (and presumably earlier), round(x) is implemented as floor(x+0.5).¹ This is a specification bug, for precisely this one pathological case.² Java 7 no longer mandates this broken implementation.³

問題

0.5+0.49999999999999994 在雙精度中正好是 1:

0.5+0.49999999999999994 is exactly 1 in double precision:

static void print(double d) {
    System.out.printf("%016x
", 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
}

這是因為 0.49999999999999994 的指數(shù)小于 0.5，所以當它們相加時，它的尾數(shù)會移動，ULP 會變大.

This is because 0.49999999999999994 has a smaller exponent than 0.5, so when they're added, its mantissa is shifted, and the ULP gets bigger.

解決方案

從 Java 7 開始，OpenJDK(例如)這樣實現(xiàn)它:⁴

Since Java 7, OpenJDK (for example) implements it thus: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;
}

<小時>

<子>1. http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#round%28double%29

<子>2. http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675(感謝@SimonNickerson 找到這個)

_{2. http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675 (credits to @SimonNickerson for finding this)}

<子>3. http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29

<子>4. http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7u40-b43/java/lang/Math.java#Math.round%28double%29

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