Do C++ signed types have modulo semantics?

[Nathan Sidwell]

Gabriel Dos Reis wrote:
> Nathan Sidwell <nathan@codesourcery.com> writes:
> 
> | Gabriel Dos Reis wrote:
> | > Michael Veksler <VEKSLER@il.ibm.com> writes:
> | > [...]
> | > | The code is not very simple, and different codes will get optimized
> | > | differently.
> | > | The user will have to learn how to write this piece of code differently for
> | > | each
> | > | processor to have best results.
> | > | | int wrap_sum(int a, int b)
> | > | {
> | > |   if ( (a<0) != (b<0))
> | > |     return a+b; // Different sign, no overflow possible.
> | > |   unsigned sum = (unsigned) a + b;
> | > |   if (sum <= MAX_INT)
> | > |     return sum;
> | > |   sum -= MIN_INT;
> | > |   if (sum > MAX_INT) // can be eliminated for 2's complement
> | > |    abort(); // oops
> | > |   return (int)sum + MIN_INT;
> | > | }
> | > | | It does not look too good optimization-wise.
> | > Thanks for providing this example.  This is the sort of thing I
> | > faced.  With pipeline process, it does harm.
> | 
> | Please explain why
> | int wrap_sum (int a, int b)
> | {
> |   return (int) ((unsigned)a + (unsigned)b));
> | }
> | is unacceptable
> 
> Notice that in your rendition you're assuming that you can convert any
> unsigned value > INT_MAX to a int without invoking undefined behaviour.
> 
> Thus the question is whether you're accepting that as documented
> behaviour of GCC, and if so, then I'm interested in which ways it
> would differs from saying that is_modulo is true.  That is useful for
> improving over the current situation, may I point out.

I give up.

nathan
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