Boost logo

Boost :

From: David Bergman (davidb_at_[hidden])
Date: 2002-08-22 13:38:42


I assume the encryption routine implementers are kind of interested in
getting "correct" results...

And, obviously, signed integer arithmetics cannot take place in a modulo
ring, since such a ring does not include negative numbers...

But, that is not a loop-hole for using double-precision in interims,
since the positive numbers of "int" (just to take an integral example)
have to be a subset of "unsigned int", so see, you still cannot maneuver
out of the modulo ring...

I.e., it is still invalid of an implementation to do the trick you


-----Original Message-----
From: boost-bounces_at_[hidden]
[mailto:boost-bounces_at_[hidden]] On Behalf Of Bill Seymour
Sent: Thursday, August 22, 2002 1:55 PM
To: boost_at_[hidden]
Subject: [boost] Re: (boost) Re: A pure C/C++ question

Victor A. Wagner wrote:
> I'm one of those weirdos that thinks that "correct" is
> a lot more important than "fast".

Perhaps in your application domain. Folk who write,
e.g., encryption routines would disagree. They need
arithmetic to be fast and repeatable (gets the same
answer every time). Whether it gets the "correct"
answer is totally irrelevant.

David Bergman wrote:
> [Integer] arithmetic is supposed to be module 2^n, ...

No, just unsigned integer arithmetic. As others have
already said, signed integer overflow yields undefined behavior. That
includes, but is not limited to, wrapping modulo 2^n as if it were
unsigned (MSVC and others), INT_MAX + 1 == 1 (Borland, IIRC), hardware
traps (Unisys, Cray, IIRC), ...

--Bill Seymour
Unsubscribe & other changes:

Boost list run by bdawes at, gregod at, cpdaniel at, john at