> but if you use them all, you'll fail to have a uniform distribution
Only if you generate them all with equal probability.
Only if you generate them all with equal probability.
Think of it as binary subdivision/search of [0,1], using a stream of random bits. Steps:
(1) Divide the current interval in 2 (using its midpoint); (2) Using one random bit, pick one of the 2 intervals; (3) if the picked interval (new current interval) lies entirely on the domain of a single floating point value[def1], stop and return this value, else go to (1).
[def1] The domain associated to a floating point value is the interval between the midpoint between it and its lower neighbor on the left, and the midpoint between it and its higher neighbor on the right.
I expect the performance is very poor, but it does cover all floating point numbers in [0,1] with exactly correct probabilities (assuming the bit probabilities are exactly correct). That's in part because naively you need higher precision arithmetic to do so, as well as up to 53 or so iterations, on average well over 32 I presume.
(I've left out the proof that probabilities are correct, but I believe it's easy to show)
The way I would define a uniform distribution is the following:
For any two floating-point numbers, r1 and r2, which form the range [r1,r2] over the real numbers, and any second pair of floating point numbers s1 and s2, which form a range [s1,s2] over the real numbers, which is contained in [r1,r2]. The probability of getting a result in [s1,s2] when sampling from [r1,r2] must be equivalent to the result of (s2-s1)/(r2-r1) with infinite precision.
This is obviously possible to achieve.