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 ge

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; }