On Thu, Jul 16, 2009 at 11:00 AM, Li Feng<nemoking...@gmail.com> wrote:
> Hi Richard,
> On 7/16/09, Richard Guenther <richard.guent...@gmail.com> wrote:
>> On Thu, Jul 16, 2009 at 1:15 AM, Tobias
>> Grosser<gros...@fim.uni-passau.de> wrote:
>>> On Wed, 2009-07-15 at 22:48 +0200, Richard Guenther wrote:
>>>> On Wed, Jul 15, 2009 at 10:46 PM, Richard
>>>> Guenther<richard.guent...@gmail.com> wrote:
>>>> > On Wed, Jul 15, 2009 at 9:15 PM, Tobias
>>>> > Grosser<gros...@fim.uni-passau.de> wrote:
>>>> >>> A note on Lis final graph algorithm. I don't understand why you
>>>> >>> want
>>>> >>> to allow data-references to be part of multiple alias-sets? (Of
>>>> >>> course
>>>> >>> I don't know how you are going to use the alias-sets ...)
>>>> >>
>>>> >> Just to pass more information to Graphite. The easiest example might
>>>> >> be
>>>> >> something like
>>>> >>
>>>> >> A -- B -- C
>>>> >>
>>>> >> if we have
>>>> >>
>>>> >> AS1 = {A,B}
>>>> >> AS2 = {B,C}
>>>> >>
>>>> >> we know that A and C do not alias and therefore do not have any
>>>> >
>>>> > No, from the above you _don't_ know that. How would you arrive
>>>> > at that conclusion?
>>>>
>>>> What I want to say is that, if A -- B -- C is supposed to be the alias
>>>> graph
>>>> resulting from querying the alias oracle for the pairs (A, B), (A, C),
>>>> (B, C)
>>>> then this is a result that will never occur. Because if (A, B) is true
>>>> and (B, C) is true then (A, C) will be true as well.
>>>
>>> What for example for this case:
>>>
>>> void foo (*b) {
>>> int *a
>>> int *c
>>>
>>> if (bar())
>>> a = b;
>>> else
>>> c = b;
>>> }
>>>
>>> I thought this may give us the example above, but it seems I am wrong.
>>> If the alias oracle is transitive that would simplify the algorithm a
>>> lot. Can we rely on the transitivity?
>>
>> Actually I was too fast (or rather it was too late), an example with
>> A -- B -- C would be
>>
>> int a, c;
>> void foo(int *p)
>>
>> with B == (*p). B may alias a and c but a may not alias c.
>>
>> So, back to my first question then, which is still valid.
>>
>> Just to pass more information to Graphite. The easiest example might be
>> something like
>>
>> A -- B -- C
>>
>> if we have
>>
>> AS1 = {A,B}
>> AS2 = {B,C}
>>
>> we know that A and C do not alias and therefore do not have any
>> dependencies.
>>
>> How do you derive at 'A and C do not alias' from looking at
>> the alias set numbers for AS1 and AS2. How do you still
>> figure out that B aliases A and C just from looking at
>> the alias set numbers? Or rather, what single alias set number
>> does B get?
> AS1 = {A,B}
> AS2 = {B,C}
>
> B is not neccessary to have only a single alias set number,
> for this situation, B will have alias number both 1 and 2 (it
> is in both alias set),
> A will be with alias number 1 and
> C will be with alias number 2.
> So A and C got different alias set number, we could conclude
> that they are not alias.
> While for A and B or B and C, as B got alias number both 1 and 2,
> so they may alias.
I see. That would work.
Richard.
> Li
>
