Java中对NaN的困惑

Question

int i = 0, j = 0;
double nan1 = (double)0/0;
double nan2 = (double)0/0;
double nan3 = (double)i/j;
System.out.println(Double.doubleToRawLongBits(nan1) == Double.doubleToRawLongBits(nan2));
System.out.println(Double.doubleToRawLongBits(nan1) == Double.doubleToRawLongBits((double)0/0));
System.out.println(Double.doubleToRawLongBits(nan3) == Double.doubleToRawLongBits(nan2));

输出:

true
true
false

请帮助我输出true前两个和false最后一个的输出.请告诉我Double.doubleToRawLongBits()方法的实际工作是什么.

Answer 1

请尝试运行以下代码来查看这些值：

public class Test
{
    public static void main(String[] args){
        int i = 0, j = 0;
        double nan1 = (double)0/0;
        double nan2 = (double)0/0;
        double nan3 = (double)i/j;
        System.out.println(Double.doubleToRawLongBits(nan1) + " == "+ Double.doubleToRawLongBits(nan2) + " is " +
            (Double.doubleToRawLongBits(nan1) == Double.doubleToRawLongBits(nan2)));
        System.out.println(Double.doubleToRawLongBits(nan1) + " == "+ Double.doubleToRawLongBits((double)0/0) + " is " +
            (Double.doubleToRawLongBits(nan1) == Double.doubleToRawLongBits((double)0/0)));
        System.out.println(Double.doubleToRawLongBits(nan3) + " == "+ Double.doubleToRawLongBits(nan2) + " is " +
            (Double.doubleToRawLongBits(nan3) == Double.doubleToRawLongBits(nan2)));
    }
}

在我的 Mac 上，它会产生以下输出：

9221120237041090560 == 9221120237041090560 is true
9221120237041090560 == 9221120237041090560 is true
-2251799813685248 == 9221120237041090560 is false

doubleToRawLongBits 方法的 Javadoc 中记录了此陷阱：

如果参数为 NaN，则结果是表示实际 NaN 值的长整数。与 doubleToLongBits 方法不同，doubleToRawLongBits 不会将所有将 NaN 编码为单个“规范”NaN 值的位模式。