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


On Sun, May 13, 2012 at 8:16 PM, Joern Rennecke <amylaar@spamcop.net> wrote:
> 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.

What is the run-time result when overflow happens?

-- 
H.J.


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