It would probably be faster to directly 'hack' a raw IEEE 754 floating number:

http://en.wikipedia.org/wiki/IEEE_754-1985

The helpful Examples section already contains the numbers 0 and 1.

Let us assume that the output is supposed to be between 0 and 1; it seems that 0 should be inclusive and 1 exclusive for the most balanced 'quick' output.

Those two numbers have the leading binary format:

0 (8 or 11 bits exponent) (23 or 52 bits fraction) For either 32 or 64 bit floats.

Thus for a float (32 bit):

0 0 (7 bits) (23 bits) 30 bits of randomness are necessary, HOWEVER 0 01111111 is already one when the fraction is all 0s. The easiest solution I can see is to evenly drop some precision:

A reference for the casting procedure: http://en.wikipedia.org/wiki/Fast_inverse_square_root#Overvi...

Pseudo:

    uint32 randomNumber;
    ... set the random somehow, maybe rand() maybe /dev/urandom etc ...
    randomNumber &= 0x3F7FFFFF;
    return * (float *) &randomNumber;

There's a really big problem with dumping random bytes straight into a float, which is that you'll end up with a distribution that, while random, is not linear - i.e. the chance of it being between 0 and 0.5 will not be the same as the chance of it being between 0.5 and 1.0. This is because the gaps between representable numbers increases as the magnitude increases, i.e. there are "more numbers" between 0 and 1 than 100 and 101. The best technique I've found (so far) is to dump the bytes into an int, and divide.

If you put 0 (i.e., plus) in the sign bit, and 1023 (i.e., zero) in the exponent field, and 52 random bits in the fraction field, of an IEEE double, then you get a number between 1 (inclusive) and 2 (exclusive); the representable numbers in that range are evenly spaced, and if your random bits are good then they are all equally likely, so you've got a uniform random number between 1 and 2. Subtract 1 if you want a uniform random number between 0 and 1 (this doesn't lose information; if a number between 1 and 2 is representable, subtracting 1 yields a representable number).

If your machine happens to work the same way as mine, the following will do it (unportable, untested, yadda yadda):

    double randomDouble() {
      uint64_t bits = random64bits();
      bits &= ~(3ull<<62); // clear sign and MSB of expt
      bits |= (1023ull<<52); // rest of exponent is 1-bits
      return (*(double*)&bits) - 1;
    }

Interesting, I'll test this. Thanks!

Btw, what is the advantage of this over dividing? Is it faster, does it give better distribution?

It should be faster than dividing. It will give you a uniform distribution over some (but not all) of the representable values between 0.0 and 1.0 (there are as many representable values between e.g. 0.25 and 0.5 as there are between 0.5 and 1.0, so a uniform distribution over representable values probably isn't what you want), which is probably as good as you can do.

Now that you mention it, that would also be an issue for the technique you describe; specifiable to the rounding characteristics for truncated numbers.

It's probably a better design idea to have a predictable sequence of float (like, application dependent) results and randomly select from within said sequence.

In my (naive and possibly flawed) tests, the division method is giving uniformly distributed results. Can you explain more what the limitations of that approach are?

The default rounding method is round-to-even, so e.g. 0.9999999 would be expected to come up slightly more often than 0.99999994 (the last two floats smaller than 1).

Being more precise would probably require thinking about all the possible int values.

I'm dividing by UULONG_MAX + 1.0, because it expects a number between 0 (inclusive) and 1.0 (exclusive). I'm not sure how portable it is though.