通过MPI进程同步数组，如果每个线程改变了一部分? [英] Synchronize array over MPI processes, if each thread changed a part of it?
问题描述
我有一个要使用 MPI 并行化的程序.我之前没有与 MPI 合作过.
I have a program I want to parallelize using MPI. I have not worked with MPI before.
该程序计算大量对象随时间的行为.的数据这些对象存储在数组中，例如double precision :: body_x(10000) x 坐标.
The program calculates the behavior for a large numer of objects over time. The data of
these objects is stored in arrays, e.g. double precision :: body_x(10000) for the x coordinate.
要计算对象的行为，需要有关所有其他对象的信息，所以每个线程都需要保存所有数据，但只会更新其中的一部分.但在此之前每个线程需要从所有其他线程获取信息的新时间步.
To calculate the behavior of an object the information about all other objects is needed, so every thread needs to hold all data but will only update a portion of it. But before the new timestep every thread needs to get the information from all other threads.
据我所知 MPI_Allgather 可以用于此目的，但它需要一个发送缓冲区和一个接收缓冲区.如果每个线程更新，如何在不同线程上同步数组数组的不同部分?我是否必须将整个数组从每个线程发送到master 在接收缓冲区中，更新 masters 数组的特定部分，毕竟线程是否已从主站重新广播其数据?
As I understand MPI_Allgather could be used for this, but it needs a send buffer and a
recive buffer. How can I synchronize an array over different threads if each thread updated
a different part of the array? Do I have to send the whole array from each thread to the
master in a recive buffer, update the specific part of the masters array and after all
threads have sent their data re-broadcast from the master?
这是一个非常基本的问题，但我对 MPI 很陌生，我发现的所有示例都是非常简单，不涉及这一点.感谢您的帮助.
This is a pretty basic question, but I'm very new to MPI and all examples I found are pretty simple and do not cover this. Thanks for any help.
伪示例(假设 Fortran 样式向量的第一个索引为 1):(是的，发送/接收最好是非阻塞的，这是为了简单起见)
Pseudo-Example (assuming Fortran-Style vectors with first index 1): (Yes the send/recive would better be done non-blocking, this is for the sake of simplicity)
if (master) then
readInputFile
end if
MPI_Bcast(numberOfObject)
allocate body_arrays(numberOfObjects)
if (master) then
fill body_arrays ! with the data from the input file
end if
MPI_Bcast(body_arrays)
objectsPerThread = numberOfObjects / threadCount
myStart = threadID * objectsPerThread + 1
myEnd = (threadID + 1) * objectsPerThread
do while (t < t_end)
do i = myStart, myEnd
do stuff for body_arrays(i)
end do
! here is the question
if (.not. master)
MPI_Send(body_arrays, toMaster)
else
do i = 1, threadCount - 1
MPI_Recive(body_arrays_recive, senderID)
body_arrays(senderID*objectsPerThread+1, (senderId+1)*objectsPerThread) = body_arrays_recive(senderID*objectsPerThread+1, (senderId+1)*objectsPerThread)
end if
MPI_Bcast(body_arrays)
! ----
t = t + dt
end do
推荐答案
听起来您想要 MPI_Allgather.为了避免需要单独的发送缓冲区，您可以使用 MPI_IN_PLACE 值.这告诉 MPI 对发送和接收使用相同的缓冲区.
It sounds like you want MPI_Allgather. To avoid needing a separate send buffer, you may be able to use the MPI_IN_PLACE value. That tells MPI to use the same buffer for both send and receive.
参见 http://mpi-forum.org/docs/mpi-2.2/mpi22-report/node99.htm#Node99
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