Re: IEEE arithmetic handling

jlg@cochiti.lanl.gov (J. Giles) writes:
  > This is easily accomplished. Let `b' be the base of your floating point
  > representation and let `p' be the number of digits in your significand.
  > Then, output followed by input is the identity if:
  >
  > (m-1) p
  > 10 >= b - 1
  >
  > Where `m' is the number of decimal digits output.


bill@amber.csd.harris.com (Bill Leonard) writes:
  > This doesn't seem to work. Take an example: let b=2, p=4. Thus
  > p
  > b - 1 = 15, so m=3. Now consider the binary number 1.001, which has
  > 4 binary digits (as dicated by p=4). This number, in decimal, is exactly
  > 1.125, which requires 4 decimal digits, thus exceeding the number m.


No problem, you output 1.12 or 1.13 whatever you fancy. Now when you read
that back in you get your original back. Unless your read routines are
wrong of course and read 1.12 or 1.13 back in as 1.25 or 1.00. That is
the major part, not that output is exact but that output followed by input
is exact.


  > (Of course, a computer using base 3 would be pretty strange.
  > :-)


On occasion on alt.folklore.computers you will read about that Russian
ternary beast.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl


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