Mm,
to be sure, if i have one processor who does:
MPI_IBcast(MPI_COMM_WORLD, request_1) // first Bcast
MPI_IBcast(MPI_COMM_WORLD, request_2) // second Bcast
it means that i can't have another process who does the follow:
MPI_IBcast(MPI_COMM_WORLD, request_2) // firt Bcast for another process
MPI_IBcast(MPI_COMM_WORLD, request_1) // second Bcast for another process
Because first Bcast of second process matches with first Bcast of first process, and it's wrong.
Is it right?
2010/9/23 Jeff Squyres
<jsquyres@cisco.com>
On Sep 23, 2010, at 6:28 AM, Gabriele Fatigati wrote:
> i'm studing the interfaces of new collective routines in next MPI-3, and i've read that new collectives haven't any tag.
Correct.
> So all collective operations must follow the ordering rules for collective calls.
Also correct.
> From what i understand, this means that i can't use:
>
> MPI_IBcast(MPI_COMM_WORLD, request_1) // first Bcast
> MPI_IBcast(MPI_COMM_WORLD, request_2) // second Bcast
No, not quite right. You can have multiple outstanding ibcast's -- they'll just be satisfied in the same order in all participating MPI processes.
> but is it possible to do this:
>
> MPI_IBcast(MPI_COMM_WORLD, request_1) // first Bcast
> MPI_IReducet(MPI_COMM_WORLD, request_2) // othwer collective
Correct -- this is also possible.
More generally, you can have multiple outstanding non-blocking collectives on a single communicator -- it doesn't matter if they are the same or different collective operations. They will each be unique instances and will be satisfied in order.
--
Jeff Squyres
jsquyres@cisco.com
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Ing. Gabriele Fatigati
Parallel programmer
CINECA Systems & Tecnologies Department
Supercomputing Group
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