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.
Li
