Will backend ever see an memory operand with address wrap around?

[Joern Rennecke]

Quoting "H.J. Lu" <hjl.tools@gmail.com>:

> On Sun, May 13, 2012 at 6:17 PM,  <amylaar@spamcop.net> wrote:
>> Quoting "H.J. Lu" <hjl.tools@gmail.com>:
>>
>>> What is the expect run-time behavior when a + b has
>>> overflow/underflow?
>>
>>
>> The expectation is wrap-around.  Note that loop strenght reduction can
>> cause assumed wrap-around semantics in RTL for strictly conforming C input
>> where no such wrap-around is in evidence.
>
> I noticed that also.  But my impression is loop strength reduction doesn't
> use wrap-around address for load/store directly.

I've actually seen it for loop strength reduction, but here is
an example that does not even involve loop strength reduction to
get into trouble - it just involves the distributive law in the
indexed access itself:

extern int a[];

void f (int o)
{
   int i;
   for (i = C; i < C + 100; i++)
     {
        a[o-i] = 0;
     }
}

At -O2, gcc (GCC) 4.7.0 20120504 (Red Hat 4.7.0-4) for i686 gives:

f:
.LFB0:
         .cfi_startproc
         movl    4(%esp), %ecx
         movl    $100, %eax
         .p2align 4,,7
         .p2align 3
.L2:
         leal    (%eax,%ecx), %edx
         subl    $1, %eax
         movl    $0, a-1200000400(,%edx,4)
         jne     .L2
         rep
         ret
         .cfi_endproc

Now consider what happens if o == C, and a is within the first GB.
unless the base address is encoded as 64 bit, you'll have an overflow.