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NNTP-Posting-Date: Fri, 24 Jun 2011 12:32:20 -0500
From: Pete Becker <pete@versatilecoding.com>
Organization: Roundhouse Consulting, Ltd.
Newsgroups: comp.lang.c++
Date: Fri, 24 Jun 2011 13:32:20 -0400
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Subject: Re: (simple?) problem with multiplication
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On 2011-06-24 13:09:56 -0400, SG said:

> In case this hasn't been mentioned already (I just skimmed this
> thread):
> 
>   (a*b)%B = ((a%B)*(b%B))%B
> 

Yes, but it doesn't solve the problem. :-( Suppose a*b overflows and 
that B is greater than both a and b. Then a%B is a, and b%B is b; then 
(a%B)*(b%B) is a*b, and it overflows.

Another approach is brute force: write double-precision multiplication 
and division routines; not too hard, but involves tedious details;

-- 
  Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The 
Standard C++ Library Extensions: a Tutorial and Reference 
(www.petebecker.com/tr1book)