Simplification of relational expressions

[Andrew Haley]

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.