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Subject: Re: [boost] Algebraic abstractions related to (big) integers in C++
From: Dave Abrahams (dave_at_[hidden])
Date: 2011-03-27 09:48:08


On Sun, Mar 27, 2011 at 9:41 AM, Brent Spillner <spillner_at_[hidden]> wrote:
>>"What I consider especially strange from an algebraic point of view, is to automatically promote a signed integer type to its corresponding unsigned integer type, when >an arithmetic expression contains both signed and unsigned integer types. Shouldn't it be the other way round? Or wouldn't it be even better, if there would be no >automatic conversion at all? At least from an algebraic point of view, any natural number is also an integer, not vice versa like in C."
>
> To be precise, this happens only when the rank of the unsigned type is
> greater than or equal to that of the signed type.  In the case where
> both operands have non-negative values, this minimizes the risk of
> overflow during addition or multiplication.  When performing an
> operation on a unsigned value and a signed value (or subtracting a
> unsigned value from a signed positive value of lesser magnitude), it
> is incumbent upon the programmer to have some notion of what he or she
> is doing.  I believe the rationale may be that signed values are the
> default, so if you went out of your way to declare something as
> unsigned then you're accepting some responsibility for using it
> appropriately.  Note also that (at least in ANSI C) when performing
> arithmetic between an unsigned value and a signed value of higher
> rank, the promotion is to the signed type--- so the "promote to the
> 'larger' type" principle is consistently applied across the board,
> particularly if you consider unsigned types "larger" than the signed
> equivalent (seems reasonable to me given how uncommon negative values
> are in most integer code.)

Even if that's consistent, it's mathematically wrong to convert a
signed type to unsigned unless you know the value is non-negative.

-- 
Dave Abrahams
BoostPro Computing
http://www.boostpro.com

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