我正在尝试减少(求和)由 MPI_type_vector 创建的派生数据类型。当我运行代码时，它崩溃并提示减少 MPI_SUM 没有为非固有数据类型定义。我写了一段简单的代码来说明我的问题。该代码尝试减少 3*3 矩阵的对角线元素:

#include "mpi.h"
#include 

int main(int argc, char *argv[]) {
int rank, size, i, j;
double a[3][3] ;

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

MPI_Datatype diag3;
MPI_Type_vector(3,1,4,MPI_DOUBLE,&diag3);
MPI_Type_commit(&diag3);

if(rank==0)
    for(i=0; i < 3 ; i++)
       for(j=0; j < 3 ; j++)
           a[i][j]=1;

if(rank==1)
    for(i=0; i < 3 ; i++)
       for(j=0; j < 3 ; j++)
           a[i][j]=-1;

MPI_Allreduce( MPI_IN_PLACE, &a[0][0], 1, diag3, MPI_SUM, MPI_COMM_WORLD );


for(i=0; i < 3 ; i++)
       for(j=0; j < 3 ; j++)
           printf("rank=%d\ta[%d][%d]=%f\n",rank,i,j,a[i][j]);

MPI_Finalize();

}

运行后报错是这样的:

*** An error occurred in MPI_Allreduce: the reduction operation MPI_SUM is not defined for non-intrinsic datatypes
*** reported by process [140130307538945,1]
*** on communicator MPI_COMM_WORLD
*** MPI_ERR_OP: invalid reduce operation

我认为 Reduce 和 MPI_SUM 可以像 MPI 文档所说的那样在派生数据类型上执行。那么，代码中存在什么问题？

câu trả lời hay nhất

Ed Smith 是对的，您需要定义自己的操作；但对于非连续类型，它需要比他列出的版本更复杂一些。下面我们有一个 add_double_vector 函数，它将解码任何 double_vector 类型并对其进行操作；它相对直接地扩展到 len > 1。

#include "mpi.h"
#include 

void add_double_vector(void *in, void *inout, int *len, MPI_Datatype *dtype)
{
    double *invec = in;
    double *inoutvec = inout;
    int nints, naddresses, ntypes;
    int combiner;

    if (*len != 1) {
        fprintf(stderr,"my_add: len>1 not implemented.\n");
        trở lại;
    } 

    MPI_Type_get_envelope(*dtype, &nints, &naddresses, &ntypes, &combiner); 
    if (combiner != MPI_COMBINER_VECTOR) {
        fprintf(stderr,"my_add: do not understand composite datatype.\n");
        trở lại;
    } 

    int vecargs [nints];
    MPI_Aint vecaddrs[naddresses];
    MPI_Datatype vectypes[ntypes];

    MPI_Type_get_contents(*dtype, nints, naddresses, ntypes, 
            vecargs, vecaddrs, vectypes);

    if (vectypes[0] != MPI_DOUBLE) {
        fprintf(stderr,"my_add: not a vector of DOUBLEs.\n");
    }

    int count = vecargs[0];
    int blocklen = vecargs[1];
    int stride = vecargs[2];

    for ( int i=0; i
        for ( int j=0; j
            inoutvec[i*stride+j] += invec[i*stride+j]; 
        } 
    }
}

int main(int argc, char *argv[]) {
    int rank, size, i, j;
    const int n=3;
    double a[n][n] ;

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

    MPI_Datatype diag3;
    MPI_Type_vector(n,1,n+1,MPI_DOUBLE,&diag3);
    MPI_Type_commit(&diag3);

    if(rank==0)
        for(i=0; i < n ; i++)
            for(j=0; j < n ; j++)
                a[i][j]=1;

    if(rank==1)
        for(i=0; i < n ; i++)
            for(j=0; j < n ; j++)
                a[i][j]=-1;

    MPI_Op vector_add;
    MPI_Op_create( add_double_vector, 1, &vector_add );
    MPI_Allreduce( MPI_IN_PLACE, &a[0][0], 1, diag3, vector_add, MPI_COMM_WORLD );
    MPI_Op_free( &vector_add );

    for(i=0; i < n ; i++)
        for(j=0; j < n ; j++)
            printf("rank=%d\ta[%d][%d]=%f\n",rank,i,j,a[i][j]);

    MPI_Finalize();

}

编译运行给出正确答案:

$ mpicc -o foo foo.c -std=c99
$ mpirun -np 2 ./foo 
rank=1 a[0][0]=0.000000
rank=1 a[0][1]=-1.000000
rank=1 a[0][2]=-1.000000
rank=1 a[1][0]=-1.000000
rank=1 a[1][1]=0.000000
rank=1 a[1][2]=-1.000000
rank=1 a[2][0]=-1.000000
rank=1 a[2][1]=-1.000000
rank=1 a[2][2]=0.000000
rank=0 a[0][0]=0.000000
rank=0 a[0][1]=1.000000
rank=0 a[0][2]=1.000000
rank=0 a[1][0]=1.000000
rank=0 a[1][1]=0.000000
rank=0 a[1][2]=1.000000
rank=0 a[2][0]=1.000000
rank=0 a[2][1]=1.000000
rank=0 a[2][2]=0.000000

关于c - mpi_allreduce 对派生数据类型 vector 求和，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29285883/