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From: David Bergman (davidb_at_[hidden])
Date: 20020822 12:07:02
Remember that this case of extending passed the 2^n world is totally
different from the legal optimization of a compiler "remembering" that
integer division was used and keeping the remainder from that operation
for a subsequent "%"...
Original Message
From: boostbounces_at_[hidden]
[mailto:boostbounces_at_[hidden]] On Behalf Of Victor A. Wagner,
Jr.
Sent: Thursday, August 22, 2002 11:58 AM
To: boost_at_[hidden]
Subject: Re: [boost] Re: A pure C/C++ question
At Thursday 2002/08/22 01:23, you wrote:
>Maxim Shemanarev writes:
> > > If the committee is seriously considering ways to handle stuff,
>how
> about
> > > dealing with the two results returned by almost every integer
divide
> > > instruction... quotient and remainder.
> >
> > Right. That's another thing. But in this case if you write x=a/b;
y=a%b; a
> > compiler at least *can try* to optimize this. In case of x*a/b it
cannot
> > because it's not allowed to!
>
>Everyone is forgetting that integer overflow on signed types results in
>undefined behaviour.
This is an artifact of the language definition (a poor one, IMO). When
_I_
learned multiplication (way back when I was 6 or 7) I noticed pretty
quick
that if I multiplied a n digit number by an m digit number I _might_ get
an
m+n digit number. The first several computers I worked on always
m+produced
a "two word" product from two "one word" operands. I was somewhat
disappointed to learn that the HLL's all seemed to ignore this "basic
fact"
of mathematics. <shrug> it's what we're stuck with, tho I still
consider
it mostly intellectual laziness that led to it. Especially with
strongly
typed languages now.
>Since the only case in which you need the intermediate
>result to be larger than int is where (x*a) overflows, this expression
>yields undefined behaviour in such cases. The upshot of this is that a
>compiler is permitted (as an extension) to _define_ such behaviour, and
>ensure that x*a/b always yields the mathematically correct result if
>the final result fits in an int. Obviously, if the final result doesn't
>fit in an int, you still have overflow (e.g. INT_MAX*INT_MAX/1), but
>this actually allows implementations to define the result of a*b/c as
>the mathematically correct answer where the final answer is in range.
allowing isn't the same as mandating
>Anthony
>
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Victor A. Wagner Jr. http://rudbek.com
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most dangerous words in the English language:
"There oughta be a law"
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