From: terje.mathisen@tmsw.no   
      
   MitchAlsup wrote:   
   >   
   > anton@mips.complang.tuwien.ac.at (Anton Ertl) posted:   
   >   
   >> Terje Mathisen  writes:   
   >>> In BID we would do division by 10 with a reciprocal multiplication that   
   >>> handles powers of 5, this way we (i.e Michael S) can fit 26/27 digits   
   >>> into a 64-bit int.   
   >>   
   >> 2^64 < 10^20   
   >>   
   >> How do you put 26/27 decimal digits in 64 bits.  Or do you mean 27   
   >> quinary digits?  What would that be good for?   
   >   
   > it takes 3.32 binary digits to encode 10, thus there are only 19.25   
   > decimal digits in 64-bits.   
      
   Michael's idea was to split the division by a power of ten into two   
   parts: A division by a power of 5 and a bitshift for the 2^N.   
      
   If we start with the bitshift (but remember the bits shifted out from   
   the bottom, then 5^26 fits into 2^64.   
      
   Does that make sense?   
      
   Terje   
      
   --   
   -    
   "almost all programming can be viewed as an exercise in caching"   
      
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