On 2006-12-19 10:44:25 -0800, Paul Eggert wrote:
> Sure, but that is trickier. In many cases code operates on
> types like time_t that are signed on some platforms and
> unsigned on others. It's easy for such code to test for
> overflow if you assume wraparound arithmetic, as code like
> { sum = a + b; if ((sum < a) != (b < 0)) overflow (); } is
> valid regardless of signedness. It's not so easy if you
> cannot assume wraparound arithmetic, particularly if
> performance is an issue (not the case in GNU expr, but it is
> true elsewhere).
If a has the value -2, b has the value 1 and the user expects to
get -1 regardless of signedness of b, then this is not valid.
So, does anyone assume that the compiler should behave as if a + b
in C were a + b on the integer ring just because many programmers
think so, even though this would be non conforming?
--
Vincent Lefèvre <[EMAIL PROTECTED]> - Web: <http://www.vinc17.org/>
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Work: CR INRIA - computer arithmetic / Arenaire project (LIP, ENS-Lyon)
