RE: [boost] Re: (boost) Re: A pure C/C++ question

From: Anthony Williams (anthwil_at_[hidden])
Date: 2002-08-23 03:30:21

Greg Colvin writes:
 > At 12:38 PM 8/22/2002, David Bergman wrote:
 > >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
 > >described.
 > 
 > No, it is perfectly valid.  If the product of two ints overflows all
 > bets are off.

Which means, as I said before, that a compiler can define it in specific
circumstances --- i.e. they can say "a*b/c" is valid if it yields an in-range
result, even if a*b overflows. Many (x86) compilers will provide a valid
result in this case anyway, because its the fewest instructions, and undefined
behaviour on overflow means they're allowed. If the result still doesn't fit,
they'll generate a divide-by-zero error, which is also allowed as undefined
behaviour.

Anthony