I might have missed it earlier Greg but was the actual problem that this helps with? I was intrigued by the underlying problem.
Regards, Greg Dr Greg Low 1300SQLSQL (1300 775 775) office | +61 419201410 mobile SQL Down Under | Web: https://sqldownunder.com<https://sqldownunder.com/> | About Greg: https://about.me/greg.low From: Greg Keogh via ozdotnet <ozdotnet@ozdotnet.com> Sent: Tuesday, 5 July 2022 8:37 AM To: ozDotNet <ozdotnet@ozdotnet.com> Cc: Greg Keogh <gfke...@gmail.com> Subject: Re: 53-bit double 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
