16 May 1999 15:24:34 -0400

Related articles |
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Rounding with Div and Mod operators william.rayer@virgin.net (William Rayer) (1999-05-09) |

Re: Rounding with Div and Mod operators wclodius@aol.com (1999-05-16) |

Re: Rounding with Div and Mod operators ucapjab@ucl.ac.uk (Jonathan Barker) (1999-05-16) |

Re: Rounding with Div and Mod operators nr@labrador.cs.virginia.edu (Norman Ramsey) (1999-05-16) |

Re: Rounding with Div and Mod operators guerby@acm.org (Laurent Guerby) (1999-05-16) |

Re: Rounding with Div and Mod operators anton@mips.complang.tuwien.ac.at (1999-05-16) |

Re: Rounding with Div and Mod operators Scott.Daniels@Acm.Org (Scott.David.Daniels) (1999-05-16) |

Re: Rounding with Div and Mod operators cdg@nullstone.com (Christopher Glaeser) (1999-05-16) |

Re: Rounding with Div and Mod operators johan.persson@mbox319.swipnet.se (Johan Persson) (1999-05-16) |

Re: Rounding with Div and Mod operators genew@shuswap.net (1999-05-20) |

Re: Rounding with Div and Mod operators sofkam@rpi.edu (1999-05-20) |

Re: Rounding with Div and Mod operators drh@microsoft.com (Dave Hanson) (1999-05-20) |

Re: Rounding with Div and Mod operators anton@mips.complang.tuwien.ac.at (1999-05-20) |

[7 later articles] |

From: | "Scott.David.Daniels" <Scott.Daniels@Acm.Org> |

Newsgroups: | comp.compilers |

Date: | 16 May 1999 15:24:34 -0400 |

Organization: | Compilers Central |

References: | 99-05-039 |

Keywords: | arithmetic, design |

One thing I'd mention is that each CPU manufacturer builds one of the

definitions in hardware. If you don't have a definition that matches

the CPU, your definition will penalize code where both n and d are

non- negative, but are not provably (in the compiler) so. This

difference is the only justification for a rule such as C has (not

well defined), but it is, at least to me, a powerful justification.

You might try having several operators yourself: towards zero, down,

and fastest. This would allow you to irritate most every partisan

equally.

Here's an interesting question assuming a two's complement machine:

Let MinInf = <sign bit on, all others off = most negative int = -1Q0>

(points to those other old fogeys who know what -1Q0 means). What is

-1 rem MinInf ?

William Rayer wrote:

*> Dear Newsgroup*

.... what to do about div and mod ...

*> n = (n div d) * d + (n mod d)*

.... nice table showing the problem at zero but not at integer limits.

*> I think this is because on the PC the default rounding of the Intel*

*> IDIV opcode is towards zero.*

Actually, I think the rounding of the IDIV opcode is is behavior, not

a default; I don't know how to change it to do anything else. This

could simply be lack of data on my part, but I think not.

....

*> (3) The language I'm developing also includes an "int" operator which*

*> takes a real number operand and returns an integer which is the*

*> operand rounded down to the next lowest integer (if the operand is an*

*> exact integer value it is unaltered). Thus if I define div using next*

*> lowest integer rounding, it is redundant as (n div d = int(n/d))*

*> always. Thus rounding to the lower integer is redundant...*

You might want to call this thing "Floor" (or some such name).

There is also Ceiling, (which is -Floor(-X) for non-boundary cases).

You might also want to have a towards zero version so you can have the

function with the largest range of input values.

Be aware your users may feel cheated when Int( 10 * 0.1 ) = 9.

-Scott David Daniels

Scott.Daniels@Acm.Org

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