Well,
With openmpi compiled with Fortran default integer*8, MPI_TYPE_2INTEGER seem to have an incorrect size. The attached Fortran program shows it,
When run on openmpi with integer*8
Size of MPI_INTEGER is 8
Size of MPI_INTEGER4 is 4
Size of MPI_INTEGER8 is 8
Size of MPI_2INTEGER is 8 <-- Should be 16
When run on "normal" openmpi
Size of MPI_INTEGER is 4
Size of MPI_INTEGER4 is 4
Size of MPI_INTEGER8 is 8
Size of MPI_2INTEGER is 8
Yves Secretan
***@ete.inrs.ca
Avant d'imprimer, pensez à l'environnement
________________________________________
De : users-***@open-mpi.org [users-***@open-mpi.org] de la part de William Au [***@hotmail.com]
Date d'envoi : 29 juin 2012 19:15
À : ***@open-mpi.org
Objet : Re: [OMPI users] fortran program with integer kind=8 using openmpi
My concern is how do the C side know fortran integer using 8 bytes?
My valgrind check show something like:
==8482== Invalid read of size 8
==8482== at 0x5F4A50E: ompi_op_base_minloc_2integer (op_base_functions.c:631)
==8482== by 0xBF70DD1: ompi_coll_tuned_allreduce_intra_recursivedoubling (op.h:498)
==8482== by 0x5F031CB: PMPI_Allreduce (pallreduce.c:105)
==8482== by 0x62E2F22: PMPI_ALLREDUCE (pallreduce_f.c:77)
==8482== by 0x5C8934: mumps_276_ (mumps_part9.F:4667)
==8482== by 0x54D89A: dmumps_ (dmumps_part1.F:157)
==8482== by 0x43D358: dmumps_f77_ (dmumps_part3.F:6651)
==8482== by 0x41420C: dmumps_c (mumps_c.c:422)
==8482== by 0x412CB4: main (my_cExample_client.c:80)
==8482== Address 0x7369608 is 0 bytes after a block of size 8 alloc'd
==8482== at 0x4A0610C: malloc (vg_replace_malloc.c:195)
==8482== by 0xBF709B9: ompi_coll_tuned_allreduce_intra_recursivedoubling (coll_tuned_allreduce.c:158)
==8482== by 0x5F031CB: PMPI_Allreduce (pallreduce.c:105)
==8482== by 0x62E2F22: PMPI_ALLREDUCE (pallreduce_f.c:77)
==8482== by 0x5C8934: mumps_276_ (mumps_part9.F:4667)
==8482== by 0x54D89A: dmumps_ (dmumps_part1.F:157)
==8482== by 0x43D358: dmumps_f77_ (dmumps_part3.F:6651)
==8482== by 0x41420C: dmumps_c (mumps_c.c:422)
==8482== by 0x412CB4: main (my_cExample_client.c:80)
The fortran side:
INTEGER IN( 2 ), OUT( 2 )
CALL MPI_ALLREDUCE( IN, OUT, 1, MPI_2INTEGER, MPI_MINLOC,
& COMM, IERR)
The compiler options will take care of IN be INTEGER*8, but will
it do the same for MPI_2INTEGER in the C side
Thanks.
Regards,
William
Date: Fri, 29 Jun 2012 07:03:18 -0400
From: Jeff Squyres <***@cisco.com>
Subject: Re: [OMPI users] fortran program with integer kind=8 using
openmpi
To: <***@atmos.washington.edu>, Open MPI Users <***@open-mpi.org>
Message-ID: <6FFEA644-3F39-4B6E-ADD6-***@cisco.com>
Content-Type: text/plain; charset=iso-8859-1
You should not have to recompile openmpi, but you do have to use the correct type. You can check the size of integers in your fortrana nd use MPI_INTEGER4 or MPI_INTEGER8 depending on what you get.
If you configure ompi with -fdefault-integer-8, then OMPI will assume that Fortran integers are always 8 bytes, so be sure to also compile all of your MPI applications the same way. Indeed, you may want to configure OMPI with something like:
./configure FCFLAGS=-fdefault-integer-8 FFLAGS=-fdefault-integer-8 \
--with-wrapper-fflags=-fdefault-integer-8 \
--with-wrapper-fcflags=-fdefault-integer-8
This will add -fdefault-integer-8 to the mpif77 and mpif90 command lines automatically so that you *can't* compile without that flag.
Be aware that 8-byte Fortran integers *should work* in Open MPI, but it is probably not well tested. You may well run into some issues; be sure to let us know if you run into bugs. Sending small test programs that show the problem are usually the best way to help us identify/fix the precise problem.
in gfortran use
integer i
if(sizeof(i) .eq. 8) then
mpi_int_type=MPI_INTEGER8
else
mpi_int_type=MPI_INTEGER4
endif
I don't think that this should be necessary -- as long as you configured OMPI with the 8-byte-integer setting, then MPI_INTEGER should represent an 8 byte integer.
--
Jeff Squyres
***@cisco.com
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