Why does IEEE 754 reserve so many NaN values?

问题:

It seems that the IEEE 754 standard defines 16,777,214 32-bit floating point values as NaNs, or 0.4% of all possible values.

I wonder what is the rationale for reserving so many useful values, while only 2 ones essentially needed: one for signaling and one for quiet NaN.

Sorry if this question is trivial, I couldn't find any explanation on the internet.

回答1:

The IEEE-754 standard defines a NaN as a number with all ones in the exponent, and a non-zero significand. The highest-order bit in the significand specifies whether the NaN is a signaling or quiet one. The remaining bits of the significand form what is referred to as the payload of the NaN.

Whenever one of the operands of an operation is a NaN, the result is a NaN, and the payload of the result equals the payload of one of the NaN operands. Payload preservation is important for efficiency in scientific computing, and at least one company has proposed using NaN payloads for proprietary uses.

In more basic terms, a NaN doesn't carry any useful numerical information, and the entire 32 bits must be reserved anyway, so the unused bits in the significand would be otherwise wasted if there were not a payload defined in the standard.

回答2:

There is likewise a payload for 64 bit floating point numbers as well, with ~10^15 possible values. Unfortunately, implementations diverge as to how the payload should be transferred between 32 and 64 bit floating point numbers and back again, i.e. whether you preserve the most significant or least significant bits. Since payload treatment is machine specific, you need different code to deal with payloads on different machines.

I wouldn't worry too much about which NaN payload is propagated after a binary operation. NaNs are exceptional values that occur with low probability, and the probability of getting 2 of them is unlikely.