Parellel编程:如何应用'MPI_Gather'或'MPI_Scatter'而不是'MPI_Isend'和'MPI_Recv'

时间:2017-10-11 20:27:29

标签: c parallel-processing mpi scatter

这是我的代码计算素数,我想只使用这些过程之间的集体交流。但是,当我使用'MPI_Scatter'和'MPI_Gather'而不是MPI_Recv和MPI_Isend更改我的代码时出错。我应该改变什么呢?

这是我的原始代码:

MPI_Request req;
MPI_Status status;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD, &p);
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);

if( my_rank == 0){
    printf("Input(50-1000) : ");
    fflush(stdout);
    scanf("%d",&w);
    for(i=0; i<w; i++) data[i] = i+1;
}

MPI_Bcast(&w, 1, MPI_INT,0,MPI_COMM_WORLD);
MPI_Bcast(data, w, MPI_INT,0,MPI_COMM_WORLD);

if( my_rank != 0){
    x = w/(p-1);
    low = (my_rank-1)*x;
    high = low+x-1;
    for(num = data[low]; num <= data[high];num++){
        result = 0;
        t=1;
        while(num>=t){
                if(num%t==0)
                result = result +1;
            t += 1;
        }
        if(result==2) i += 1;
    }
    MPI_Isend(&i,1,MPI_INT,0,0,MPI_COMM_WORLD,&req);
}

if(my_rank == 0){
    int j = 0;
    for( j = 1; j < p; j++){
        MPI_Recv(&i,1,MPI_DOUBLE,MPI_ANY_SOURCE,0,MPI_COMM_WORLD,&status);
    printf("Process %d : There are %d prime numbers\n",status.MPI_SOURCE,i );
    }
}
MPI_Finalize();

}

输出是:

Input(50-1000) : 50
Process 1 : There are 5 prime numbers
Process 2 : There are 4 prime numbers
Process 3 : There are 2 prime numbers
Process 4 : There are 4 prime numbers

以下是我更改代码的部分:

if( my_rank != 0){
        x = w/(p-1);
        low = (my_rank-1)*x;
        high = low+x-1;
        for(num = data_send[low]; num <= data[high];num++){
            result = 0;
            t=1;
            while(num>=t){
                if(num%t==0)
                    result = result +1;
                    t += 1;
                }
            if(result==2) i += 1;
     }
 MPI_Scatter(data_send,1,MPI_INT,data_recv,1,MPI_INT,0,MPI_COMM_WORLD);
    }

2 个答案:

答案 0 :(得分:1)

MPI_Scatter是一项集体行动,需要所有职级召集。同样适用于MPI_Gather

表示:MPI_Scatter调用应移至代码中的if块之外(类似于MPI_Gather)。

您可以找到示例here

答案 1 :(得分:1)

以下是您如何使用MPI_Gather()

int * results;
if( my_rank != 0){
    x = w/(p-1);
    low = (my_rank-1)*x;
    high = low+x-1;
    for(num = data[low]; num <= data[high];num++){
        result = 0;
        t=1;
        while(num>=t){
                if(num%t==0)
                result = result +1;
            t += 1;
        }
        if(result==2) i += 1;
    }
} else {
    results = (int *)malloc(p * sizeof(int));
}

MPI_Gather(&i, 1, MPI_INT, results, 1, MPI_INT, 0, MPI_COMM_WORLD);

if(my_rank == 0){
    int j = 0;
    for( j = 1; j < p; j++){
        printf("Process %d : There are %d prime numbers\n",j, results[j]);
    }
    free(results);
}

请注意,您可以安排此算法以便在0级进行计算,因此使用p MPI任务而不是p-1