[mpich-discuss] mpich2 with ifort and -i8 -r8

[Rajeev Thakur]

MPICH2 will use what the compiler provides. If both C and Fortran compilers provide 8 byte ints consistently, you will get 8 byte ints and can send large messages. MPICH2 source will have to be compiled with those compiler options though.

Rajeev

On Feb 4, 2011, at 2:41 PM, Gus Correa wrote:

> Rajeev Thakur wrote:
>> I think you will need to pass the same options to the Intel C compiler if it can be done.
>> Rajeev
>> On Feb 4, 2011, at 1:08 PM, Pierre-Olivier Dallaire wrote:
>>> Hi,
>>> 
>>> I'm trying to compile mpich2 with Fortran support and also with options "-i8 -r8" for the
>>> ifort compiler (intel) / however, I'm getting this :
>>> 
>>> configure: error: Fortran integers and C ints are not the same size.  The current Fortran binding does not support this case.  Either force the Fortran compiler to use integers of 4 bytes, or use --disable-f77 on the configure line for MPICH2.
>>> 
>>> Trying to disable f77 does not work / is there anything else I can do ? I absolutely need to convert
>>> the real and integers.
>>> 
>>> Regards,
>>> 
>>> PO
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> 
> Hi Rajeev
> 
> Just a question while you are at it.
> 
> Which size will the MPICH2 MPI_INT[EGER] type
> (and address integers, MPI_AINT) will have if one
> compiles MPICH2 with -i8?
> Four bytes?
> Eight bytes?
> Other?
> 
> I.e., can one effectively increase the MPI[CH2] send/recv buffer sizes
> and addressable memory from 4-bytes to 8-bytes by compiling MPICH2
> with -i8?
> 
> Thank you,
> Gus Correa
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