java - 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

请帮助我前两个的输出是true，最后一个的输出是false。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以查看值:

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，它产生以下输出:

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

此陷阱记录在 doubleToRawLongBits method 的 Javadoc 中:

If the argument is NaN, the result is the long integer representing the actual NaN value. Unlike the doubleToLongBits method, doubleToRawLongBits does not collapse all the bit patterns encoding a NaN to a single "canonical" NaN value.