uint u1 = *[32 random bits]*; uint u2 = *[32 random bits]*; uint a = u1 >> 5, b = u2 >> 6; return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
Can anyone explain this magic? Is this correct?
This bit trick is no longer important to decode. I was trying to figure out how to convert a UInt64 of random bits into a normalised double in the range [0,1) in such a way that all 2^53 possible floating values could be produced. Some web sites suggested simply doing this: uint u = *[64 random bits]* double d = (u >> 11) * 1.11022302462515654e-16 // * 2-53 I was suspicious that subtle rounding problems might produce a defective output range, but a quick experiment in LINQPad that fed limit values into the calculation demonstrated that a complete set of significand bit patterns was generated at the limits. This supplies good evidence that the above shift-and-multiply converts 53 of the 64 random bits into a complete possible double range. *Greg K*