Hi,
George Bosilca wrote:
> On the all-to-all collective the send and receive buffers has to be able
> to contain all the information you try to send. On this particular case,
> as you initialize the envio variable to a double I suppose it is defined
> as a double. If it's the case then the error is that the send operation
> will send more data than the amount available on the envio variable.
>
> If you want to be able to do correctly the all-to-all in your example,
> make sure the envio variable has a size at least equal to:
> tam * sizeof(byte) * NPROCS, where NPROCS is the number of procs available
> on the mpi_comm_world communicator.
I'm unfortunately not that Fortran guy. Maybe the best would be to
submit the whole function at the beginning, it's neither secret nor big:
module alltoall
use globales
implicit none
contains
subroutine All_to_all
integer,parameter :: npuntos = 24
integer,parameter :: t_max = 2**(npuntos-1)
integer siguiente,anterior,tam,rep,p_1,p_2,i,j,ndatos,rep2,o,k
double precision time1,time2,time,ov,tmin,tavg
double precision,dimension(t_max)::envio
double precision,dimension(:),allocatable::recibe
double precision,dimension(npuntos)::m_tmin,m_tavg
double precision,dimension(npuntos)::tams
rep2 = 10
tag1 = 1
tag2 = 2
rep = 3
allocate(recibe(t_max*nproc))
siguiente = my_id + 1
if (my_id == nproc -1) siguiente = 0
anterior = my_id - 1
if (my_id == 0) anterior = nproc- 1
do i = 1,npuntos
print *,'puntos',i
tam = 2**(i-1)
tmin = 1e5
tavg = 0.0d0
do j = 1,rep
envio = 8.0d0*j
call mpi_barrier(mpi_comm_world,ierr)
time1 = mpi_wtime()
do k = 1,rep2
call mpi_alltoall(envio,tam,mpi_byte,recibe,tam,mpi_byte,mpi_comm_world,ierr)
end do
call mpi_barrier(mpi_comm_world,ierr)
time2 = mpi_wtime()
time = (time2 - time1)/(rep2)
if (time < tmin) tmin = time
tavg = tavg + time
end do
m_tmin(i) = tmin
m_tavg(i) = tavg/rep
end do
call mpi_barrier(mpi_comm_world,ierr)
print *,"acaba"
if (my_id == 0) then
open (1,file='Alltoall.dat')
write (1,*) "#Prueba All to all entre todos los procesadores(",nproc,")"
write (1,*) "#Precision del reloj:",mpi_wtick()*1.0d6,"(muS)"
do i =1,npuntos
write(1,900) 2*nproc*2**(i-1),m_tmin(i),m_tavg(i)!,ov
end do
close(1)
end if
900 FORMAT(I10,F14.8,F14.8)
800 FORMAT(I10,F14.8,F14.8)
end subroutine
end module
Can you read this? (Sorry, I can't.) But the size_of envio seems to be
2**32 = 8388608 doubles, isn't it? I don't understand, why it should
depend on the number of MPI nodes, as you said.
Thanks for your help. Kind regards,
--
Frank Gruellich
HPC-Techniker
Tel.: +49 3722 528 42
Fax: +49 3722 528 15
E-Mail: frank.gruellich_at_[hidden]
MEGWARE Computer GmbH
Vertrieb und Service
Nordstrasse 19
09247 Chemnitz/Roehrsdorf
Germany
http://www.megware.com/
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