Confusion on NaN in Java

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));

output:

true
true
false

Please help me how the output came true for first two and false for last one. Please tell me what is actual work of Double.doubleToRawLongBits() method.

Answer 1

Please try to run following code to see the values:

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)));
    }
}

On my Mac, it produces following output:

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

This pitfall is documented in the Javadoc for the doubleToRawLongBits method:

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.

Answer 2

The reason is because when you divide a double variable 0 by 0 it returns NaN, so the method doesnt have a single canonical representation in binary, so it may return the binary of NaN as 7F F8 00 00 00 00 00 00 or FF F8 00 00 00 00 00 00.

Even though technically they represent the same thing, which is NaN, its different in binary representation.