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*From*: Tobias Grosser <grosser at fim dot uni-passau dot de>*To*: Richard Guenther <richard dot guenther at gmail dot com>*Cc*: gcc at gcc dot gnu dot org, Sebastian Pop <sebpop at gmail dot com>, Li Feng <nemokingdom at gmail dot com>, gcc-graphite <gcc-graphite at googlegroups dot com>*Date*: Thu, 16 Jul 2009 01:15:04 +0200*Subject*: Re: How could I get alias set information from data_reference_p*References*: <f18356030907132301n2306cd0ah3929575851d5cdff@mail.gmail.com> <cb9d34b20907140814l405e1061k816f9cdb002e9515@mail.gmail.com> <84fc9c000907140826t645135efp72d196d04ebb9095@mail.gmail.com> <cb9d34b20907140903y3a9e067g81d35a7f4ab3d58@mail.gmail.com> <cb9d34b20907140908l2f1a2eb2r1be98670ac62d86e@mail.gmail.com> <84fc9c000907141434n18721235oddafd9ff5489ff72@mail.gmail.com> <1247655619.1418.179.camel@localhost> <84fc9c000907150426w69f13cebv4a36e834d1e3c9f1@mail.gmail.com> <1247685312.87884.417.camel@localhost> <84fc9c000907151346r58239f06u318b55fe72ae0038@mail.gmail.com> <15137_1247690941_4A5E40BC_15137_586_1_84fc9c000907151348s41395cc5u6cfacb60cde78bfa@mail.gmail.com>

On Wed, 2009-07-15 at 22:48 +0200, Richard Guenther wrote: > On Wed, Jul 15, 2009 at 10:46 PM, Richard > Guenther<richard.guenther@gmail.com> wrote: > > On Wed, Jul 15, 2009 at 9:15 PM, Tobias > > Grosser<grosser@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? Tobi

**Follow-Ups**:**Re: How could I get alias set information from data_reference_p***From:*Richard Guenther

**References**:**How could I get alias set information from data_reference_p***From:*Li Feng

**Re: How could I get alias set information from data_reference_p***From:*Sebastian Pop

**Re: How could I get alias set information from data_reference_p***From:*Richard Guenther

**Re: How could I get alias set information from data_reference_p***From:*Sebastian Pop

**Re: How could I get alias set information from data_reference_p***From:*Sebastian Pop

**Re: How could I get alias set information from data_reference_p***From:*Richard Guenther

**Re: How could I get alias set information from data_reference_p***From:*Tobias Grosser

**Re: How could I get alias set information from data_reference_p***From:*Richard Guenther

**Re: How could I get alias set information from data_reference_p***From:*Tobias Grosser

**Re: How could I get alias set information from data_reference_p***From:*Richard Guenther

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