c++ - 将阵列分散到仅 worker 任务

我有一个存储数组的数组(称为sendbuff)，我想通过使用MPI::Scatter将这些数组发送到其他线程。

                        sendbuff
   #####    ###############################
p  # 0 # -> # -1 # -1 # -1 # -1 # -1 # -1 # (values)
o  #####    ###############################
s  # 1 # -> # -1 # -1 # -1 # -1 # -1 # -1 # (values)
   #####    ###############################

可以看出，sendbuff[0]拥有大小为6的数组，该数组具有6个值(均为-1)，而sendbuff[1]具有相同的内容。我想将-1的数组发送到其他线程，并将它们保存在一个名为recvbuff的数组中，该数组填充0:

        recvbuff
#########################
# 0 # 0 # 0 # 0 # 0 # 0 #
#########################

我研究了操作系统以找到答案，但找到了一些答案，但它们使用MPI_Datatype，但我想避免使用它。
为了实现此目标，我编写了以下无效的代码:

int main( int argc, char *argv[]){

    //variable innitialization
    int taskid, ntasks, buffsize, **sendbuff, *recvbuff;

    MPI::Init(argc, argv);

    taskid = MPI::COMM_WORLD.Get_rank();
    ntasks = MPI::COMM_WORLD.Get_size();

    buffsize = 6;

    //memory innitialization
    recvbuff = new int[buffsize];
    sendbuff = new int*[ntasks];
    for(int i = 0; i < ntasks; i++){
        sendbuff[i] = new int[buffsize];
    }

    //array innitialization
    for(int i = 0; i < buffsize; i++){
        recvbuff[i] = 0;
    }
    for(int i = 0; i < ntasks; i++){
        for(int j = 0; j < buffsize; j++){
            sendbuff[i][j] = -1;
        }
    }

    //communication
    MPI::COMM_WORLD.Scatter(sendbuff[0], buffsize, MPI::INT, recvbuff, buffsize,
        MPI::INT, 0);

    //output
    for(int i = 0; i < buffsize; i++){
        cout<<"Task"<<taskid<<" recvbuff["<<i<<"] = "<<recvbuff[i] << endl;
    }

    //cleaning
    for(int i = 0; i < ntasks; i++){
            delete[] sendbuff[i];
    }
    delete[] sendbuff;
    delete[] recvbuff;


    MPI::Finalize();

    return EXIT_SUCCESS;
}

使用散点图后，我希望他的recvbuff变量可以用-1值填充，但是我会混合使用-1和垃圾，如下所示:

$ mpirun -np 3 a.out
Task0 recvbuff[0] = -1
Task0 recvbuff[1] = -1
Task0 recvbuff[2] = -1
Task0 recvbuff[3] = -1
Task0 recvbuff[4] = -1
Task0 recvbuff[5] = -1
Task1 recvbuff[0] = 33
Task1 recvbuff[1] = 0
Task1 recvbuff[2] = -1
Task1 recvbuff[3] = -1
Task1 recvbuff[4] = -1
Task1 recvbuff[5] = -1
Task2 recvbuff[0] = -1
Task2 recvbuff[1] = -1
Task2 recvbuff[2] = 33
Task2 recvbuff[3] = 0
Task2 recvbuff[4] = 1768975727
Task2 recvbuff[5] = 7496543

我做错了什么？
在此先感谢Pedro。

最佳答案

分散和聚集在in this answer中进行了详细描述。 Scatter分散数据并将其分散到其他任务，但是数据必须存储在连续的内存中-MPI_Scatter无法知道它需要跟随指针，如果是，则跟踪多少指针，以及您分配sendbuff的方式:

sendbuff = new int*[ntasks];
for(int i = 0; i < ntasks; i++){
    sendbuff[i] = new int[buffsize];
}

sendbuff的不同行可能分散在整个系统内存中。如果连续分配数据，您将快要在那里了:

sendbuff = new int*[ntasks];
sendbuff[0] = new int[ntasks * 6];
for(int i = 1; i < ntasks; i++){
    sendbuff[i] = &(sendbuff[0][i*6];
}

现在您应该可以分散了，但是请注意，第0行将排在第0位；也就是说，分散传播到通信器中的所有过程。如果您仅尝试发送非零零任务，那么最简单的方法是在sendbuff中为行0保留一行伪数据，以便正常散布可以正常工作:

#include <iostream>
#include <mpi.h>

int main(int argc, char **argv)
{
    int rank, size;
    const int nelem = 6;

    MPI_Init(&argc, &argv);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    MPI_Comm_size(MPI_COMM_WORLD, &size);

    int **sendbuff = new int*[size];
    int *recvbuff  = new int[nelem];

    if (rank == 0) {
        sendbuff[0] = new int[nelem * size];
        for (int i=0; i<size; i++)
            sendbuff[i] = &(sendbuff[0][nelem*i]);

        for (int i=0; i<size; i++)
            for (int j=0; j<nelem; j++)
                sendbuff[i][j] = i-1;
    }

    MPI_Scatter(sendbuff[0], nelem, MPI_INT, recvbuff, nelem, MPI_INT, 0, MPI_COMM_WORLD);

    if (rank != 0) {
        std::cout << "Scatter: [ " << rank << "]: ";
        for (int i=0; i<nelem; i++)
            std::cout << recvbuff[i] << " ";
        std::cout << std::endl;

        for (int i=0; i<nelem; i++)
            recvbuff[i] *= recvbuff[i];
    }

    MPI_Gather(recvbuff, nelem, MPI_INT, sendbuff[0], nelem, MPI_INT, 0, MPI_COMM_WORLD);
    if (rank == 0) {
        for (int j=1; j<size; j++) {
            std::cout << "Gather: [ " << j << "]: ";
            for (int i=0; i<nelem; i++)
                    std::cout << sendbuff[j][i] << " ";
            std::cout << std::endl;
        }
    }

    delete [] recvbuff;
    if (rank == 0)
        delete [] sendbuff[0];
    delete [] sendbuff;

    MPI_Finalize();
}

请注意，我们正在分散数据， worker 正在对数字进行平方，而主数据则将其收集回来。编译并运行可以得到:

$ mpic++ -o intercomm intercomm.cxx
$ mpirun -np 4 ./intercomm
Scatter: [ 2]: 1 1 1 1 1 1
Scatter: [ 1]: 0 0 0 0 0 0
Scatter: [ 3]: 2 2 2 2 2 2
Gather: [ 1]: 0 0 0 0 0 0
Gather: [ 2]: 1 1 1 1 1 1
Gather: [ 3]: 4 4 4 4 4 4

如果您宁愿避免使用等级0的虚拟数据(也许很大)，则可以将任务分为两组，即主任务和辅助任务，并设置一个intercommunicator以允许它们之间的集体通信。这是一个简单的程序，可以执行此操作:

#include <iostream>
#include <mpi.h>

int main(int argc, char **argv)
{
    MPI_Comm   localComm;    /* intra-communicator of local sub-group */
    MPI_Comm   interComm;    /* inter-communicator */
    int masterworker;
    int rank, size;
    const int nelem = 6;
    int rootrank;

    MPI_Init(&argc, &argv);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    MPI_Comm_size(MPI_COMM_WORLD, &size);

    masterworker = (rank == 0 ? 0 : 1);
    MPI_Comm_split(MPI_COMM_WORLD, masterworker, rank, &localComm);

    if (masterworker == 0)
    {
        MPI_Intercomm_create( localComm, 0, MPI_COMM_WORLD, 1, 1, &interComm);
        rootrank = ( rank == 0 ? MPI_ROOT : MPI_PROC_NULL );
    }
    else {
        MPI_Intercomm_create( localComm, 0, MPI_COMM_WORLD, 0, 1, &interComm);
        rootrank = 0;
    }

    int **sendbuff = new int*[size-1];
    int *recvbuff  = new int[nelem];

    if (rank == 0) {

        sendbuff[0] = new int[nelem * (size-1)];
        for (int i=1; i<size-1; i++)
            sendbuff[i] = &(sendbuff[0][nelem*i]);

        for (int i=0; i<size-1; i++)
            for (int j=0; j<nelem; j++)
                sendbuff[i][j] = i;
    }

    MPI_Scatter(sendbuff[0], nelem, MPI_INT, recvbuff, nelem, MPI_INT, rootrank, interComm);

    if (masterworker == 1) {
        std::cout << "Scatter: [ " << rank << "]: ";
        for (int i=0; i<nelem; i++)
            std::cout << recvbuff[i] << " ";
        std::cout << std::endl;

        for (int i=0; i<nelem; i++)
            recvbuff[i] *= recvbuff[i];
    }

    MPI_Gather(recvbuff, nelem, MPI_INT, sendbuff[0], nelem, MPI_INT, rootrank, interComm);
    if (masterworker == 0) {
        for (int j=0; j<size-1; j++) {
            std::cout << "Gather: [ " << j << "]: ";
            for (int i=0; i<nelem; i++)
                    std::cout << sendbuff[j][i] << " ";
            std::cout << std::endl;
        }
    }




    MPI_Comm_free(&interComm);
    MPI_Comm_free(&localComm);
    delete [] recvbuff;
    if (rank == 0)
        delete [] sendbuff[0];
    delete [] sendbuff;

    MPI_Finalize();
}

再次，编译和运行给出:

$ mpic++ -o intercomm intercomm.cxx
$ mpirun -np 4 ./intercomm
Scatter: [ 1]: 0 0 0 0 0 0
Scatter: [ 2]: 1 1 1 1 1 1
Scatter: [ 3]: 2 2 2 2 2 2
Gather: [ 0]: 0 0 0 0 0 0
Gather: [ 1]: 1 1 1 1 1 1
Gather: [ 2]: 4 4 4 4 4 4

或者，如果您不想弄乱内部通信器，只需在sendbuff中将一行虚拟数据保留为等级0，以使正常散点正常工作:

#include <iostream>
#include <mpi.h>

int main(int argc, char **argv)
{
    int rank, size;
    const int nelem = 6;

    MPI_Init(&argc, &argv);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    MPI_Comm_size(MPI_COMM_WORLD, &size);

    int **sendbuff = new int*[size];
    int *recvbuff  = new int[nelem];

    if (rank == 0) {
        sendbuff[0] = new int[nelem * size];
        for (int i=0; i<size; i++)
            sendbuff[i] = &(sendbuff[0][nelem*i]);

        for (int i=0; i<size; i++)
            for (int j=0; j<nelem; j++)
                sendbuff[i][j] = i-1;
    }

    MPI_Scatter(sendbuff[0], nelem, MPI_INT, recvbuff, nelem, MPI_INT, 0, MPI_COMM_WORLD);

    if (rank != 0) {
        std::cout << "Scatter: [ " << rank << "]: ";
        for (int i=0; i<nelem; i++)
            std::cout << recvbuff[i] << " ";
        std::cout << std::endl;

        for (int i=0; i<nelem; i++)
            recvbuff[i] *= recvbuff[i];
    }

    MPI_Gather(recvbuff, nelem, MPI_INT, sendbuff[0], nelem, MPI_INT, 0, MPI_COMM_WORLD);
    if (rank == 0) {
        for (int j=1; j<size; j++) {
            std::cout << "Gather: [ " << j << "]: ";
            for (int i=0; i<nelem; i++)
                    std::cout << sendbuff[j][i] << " ";
            std::cout << std::endl;
        }
    }

    delete [] recvbuff;
    if (rank == 0)
        delete [] sendbuff[0];
    delete [] sendbuff;

    MPI_Finalize();
}

再次编译并运行给出:

$ mpic++ -o intercomm intercomm.cxx
$ mpirun -np 4 ./intercomm
Scatter: [ 2]: 1 1 1 1 1 1
Scatter: [ 1]: 0 0 0 0 0 0
Scatter: [ 3]: 2 2 2 2 2 2
Gather: [ 1]: 0 0 0 0 0 0
Gather: [ 2]: 1 1 1 1 1 1
Gather: [ 3]: 4 4 4 4 4 4