Chris Thiel wrote:
> I have a question about some relational expressions being transformed
> into constants when neither wrapping nor strict signed overflow is used.
> Here is the example I was looking at:
>
> extern int get_int (void);
> extern void foo (int);
>
> int
> main (void)
> {
> int a = get_int ();
> foo (a + 1 > a);
> foo (a + 1 >= a);
> foo (a + 1 == a);
> foo (a + 1 <= a);
> foo (a + 1 < a);
> foo (a + 1 != a);
> return 0;
> }
>
> I was under the impression that none of these would be simplified into
> constants when using `-fno-wrapv -fno-strict-overflow'. But this is
> what is dumped into 003t.original with the current trunk.
>
> {
> int a = get_int ();
> foo (a + 1 > a);
> foo (a + 1 >= a);
> foo (0);
> foo (a + 1 <= a);
> foo (a + 1 < a);
> foo (1);
> return 0;
> }
>
> Is there a reason that the equality and inequality expressions (and only
> those expressions) are simplified? I understand that signed integer
> overflow is undefined.
I'm more than a little mystified by this question. Unless the target is
using something truly weird like saturating arithmetic we can always
guarantee that a+1 != a, but we cannot guarantee that a+1 > a.
Andrew.
