Re: [boost] Re: A pure C/C++ question

From: Victor A. Wagner, Jr. (vawjr_at_[hidden])
Date: 2002-08-22 10:57:30

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 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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