[David]

What is it, 2**48-2 signaling NaNs and 2**48 quiet NaNs? Is my quick count correct (in 64-bit)?

Any bit pattern where the exponent is all ones (there are 11 exponent bits, giving 2**(64-11) = 2**53 bit patterns with an all-ones exponent), _and_ the significand isn't all 0 (it's an infinity if they're all 0). So that's 2**53 - 2 NaN bit patterns in all. The original standard didn't define the encoding difference between signaling and quiet, so all that can be said is that there are 2**52 - 2 of one and 2**52 of the other. But in real life, almost all implementations take the most-significant bit of the significand to mean "quiet", so in almost all implementations there are 2 more qNaNs than sNaNs.

Great opportunity for steganography, I reckon.

Too easy ;-) When I send messages to my Russian handlers, I send a long list of floating-point + - * / and sqrt examples. The CIA hasn't yet figured out the secret: an example means "1" if the 56th most significant bit of the infinitely precise result is 1, else 0. Staring at the actual double precision results is useless :-)