Fri, 20 Nov 1992 16:09:35 GMT

Related articles |
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[3 earlier articles] |

Re: IEEE arithmetic handling bill@amber.csd.harris.com (1992-11-16) |

Re: IEEE arithmetic handling jlg@cochiti.lanl.gov (1992-11-17) |

Re: IEEE arithmetic handling eggert@twinsun.com (1992-11-17) |

Re: IEEE arithmetic handling tmb@arolla.idiap.ch (1992-11-18) |

Re: IEEE arithmetic handling bart@cs.uoregon.edu (1992-11-19) |

Re: IEEE arithmetic handling bill@amber.csd.harris.com (1992-11-20) |

Re: IEEE arithmetic handling bill@amber.csd.harris.com (1992-11-20) |

Re: IEEE arithmetic handling Dik.Winter@cwi.nl (1992-11-23) |

Re: IEEE arithmetic handling bright@nazgul.uucp (Walter Bright) (1993-01-07) |

Newsgroups: | comp.compilers |

From: | bill@amber.csd.harris.com (Bill Leonard) |

Organization: | Harris CSD, Ft. Lauderdale, FL |

Date: | Fri, 20 Nov 1992 16:09:35 GMT |

Keywords: | arithmetic |

References: | 92-11-041 92-11-095 |

*> 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.*

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.

I don't believe any formula like the above can work if b contains a factor

that is not also a factor of 10 -- for instance, a b of 3 would never

work, since there are fractions in base 3 that are non-terminating

decimals. (Of course, a computer using base 3 would be pretty strange.

*:-)*

--

Bill Leonard

Harris Computer Systems Division

2101 W. Cypress Creek Road

Fort Lauderdale, FL 33309

bill@ssd.csd.harris.com

--

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