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.doub         


        

Answer 1

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.