From: David Bergman (davidb_at_[hidden])
Date: 2002-08-23 11:19:55
As I wrote in another post,
Those restrictions on the "unsigned int" together with positive "int"
being a subset of "unsigned int" do give the implied invalidity.
If someone founds this argument to be illogical, please advice me.
And, yes, I am fully aware of the clause in The International Standard
saying that "all bets are off as soon as the program contains at least
one undefined operation"...
[mailto:boost-bounces_at_[hidden]] On Behalf Of Anthony Williams
Sent: Friday, August 23, 2002 4:19 AM
Subject: RE: [boost] Re: A pure C/C++ question
David Bergman writes:
> An implementation is not even allowed to do use that double-precision
> "interim". >
> Since the arithmetic is supposed to be module 2^n, where n is the
number > of bits in the integral type in question. >
> Remember that "mathematically correct" should actually be
> "mathematically correct in the module 2^n ring".
Only for unsigned types
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