为什么Math.round(0.49999999999999994)返回1 [英] Why does Math.round(0.49999999999999994) return 1
问题描述
在以下程序中,您可以看到,对于每个值略小于 .5
的四舍五入,除了 0.5
。
In the following program you can see that for each value slightly less that .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.
推荐答案
摘要
在Java 6(可能早期)中, round(x)
被实现为 floor (x + 0.5)
。 1 这是一个规范错误,正是这一个病理情况。 2 Java 7不再要求这个破坏的实现 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
问题
0.5+ 0.49999999999999994正好是双精度1:
0.5+0.49999999999999994 is exactly 1 in double precision:
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
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.
解决方案
由于Java 7,OpenJDK(例如)实现了它: 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;
}
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 (以@SimonNickerson为单位查询)
2. http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6430675 (credits to @SimonNickerson for finding this)
3. http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#round%28double%29
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