What are the other NaN values?

2018-06-20 19:25:58

The documentation for java.lang.Double.NaN says that it is

A constant holding a Not-a-Number (NaN) value of type double . It is equivalent to the value returned by Double.longBitsToDouble(0x7ff8000000000000L) .

This seems to imply there are others. If so, how do I get hold of them, and can this be done portably?

To be clear, I would like to find the double values x such that

Double.doubleToRawLongBits(x) != Double.doubleToRawLongBits(Double.NaN)

and

Double.isNaN(x)

are both true.

You need doubleToRawLongBits rather than doubleToLongBits .

doubleToRawLongBits extracts the actual binary representation. doubleToLongBits doesn't, it converts all NaN s to the default NaN first.

double n = Double.longBitsToDouble(0x7ff8000000000000L); // default NaN
double n2 = Double.longBitsToDouble(0x7ff8000000000100L); // also a NaN, but M != 0

System.out.printf("%Xn", Double.doubleToLongBits(n));
System.out.printf("%Xn", Double.doubleToRawLongBits(n));
System.out.printf("%Xn", Double.doubleToLongBits(n2));
System.out.printf("%Xn", Double.doubleToRawLongBits(n2));

output:

7FF8000000000000
7FF8000000000000
7FF8000000000000
7FF8000000000100

Java uses IEEE 754 for its floating point numbers and therefore follows their rules.

According to the Wikipedia page on NaN it is defined like this:

A bit-wise example of a IEEE floating-point standard single precision NaN: x111 1111 1axx xxxx xxxx xxxx xxxx xxxx where x means don't care.

So there are quite a few bit-patterns all of which are NaN values.

IEEE 754 defines a NaN as a number with all exponent bits which are 1 and a non zero number in the mantissa.

So for a single-precision number you are looking for:

S     E            M
x  11111111   xxxxxx....xxx (with M != 0)

Java handles this like so:

Double n = Double.longBitsToDouble(0x7ff8000000000000L); // default NaN
Double n2 = Double.longBitsToDouble(0x7ff8000000000100L); // also a NaN, but M != 0

System.out.println(n.isNaN()); // true
System.out.println(n2.isNaN()); // true
System.out.println(n2 != Double.doubleToLongBits(Double.NaN)); // true

To sum, you can use any NaN you want which conforms to the rules aforementioned (all bits 1 in exponent and mantissa != 0).

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