学习MPI

 

 

1. MPI_Reduce函数

int MPI_Reduce(void *sendBuf, void *receiveBuf, int count, MPI_Datatype dataType, 

                               MPI_Op operator, int root, MPI_Comm comm)

每个进程从sendBuf向root进程的receiveBuf发数据，通过opration函数(例如MPI_SUM)来汇总数据。 

2. 图型示例

3. 举例

 

#include<stdio.h>
#include"mpi.h"
#define SIZE 4

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

        float sendBuf[SIZE][SIZE] = {
                {1.0,   2.0,    3.0,    4.0},
                {5.0,   6.0,    7.0,    8.0},
                {9.0,   10.0,   11.0,   12.0},
                {13.0,  14.0,   15.0,   16.0}
        };

        MPI_Init(&argc, &argv);
        MPI_Comm_rank(MPI_COMM_WORLD, &rankID);
        MPI_Comm_size(MPI_COMM_WORLD, &totalNumTasks);

        float totalSum;
        if(totalNumTasks == SIZE){
                int source = 0;
                int sendCount = SIZE;
                int recvCount = SIZE;
                float recvBuf[SIZE];
                //scatter data from source process to all processes in MPI_COMM_WORLD
                MPI_Scatter(sendBuf, sendCount, MPI_FLOAT,
                            recvBuf, recvCount, MPI_FLOAT, source, MPI_COMM_WORLD);

                float sumPerProcess = recvBuf[0] + recvBuf[1] + recvBuf[2] + recvBuf[3];
                printf("my rankID = %d, receive Results: %f %f %f %f, totalOnMe = %f\n",
                        rankID, recvBuf[0], recvBuf[1], recvBuf[2], recvBuf[3], sumPerProcess);

                int count = 1;
                int root = 0;
                MPI_Reduce(&sumPerProcess, &totalSum, count, MPI_FLOAT, MPI_SUM, root, MPI_COMM_WORLD);
                printf("Process %d is done.\n", rankID);
                fflush(stdout);
        }
        else
                printf("Please specify -n %d\n", SIZE);

        MPI_Finalize();
        if(rankID == 0) printf("totalSum = %f\n", totalSum);
        return 0;
}

 

4. 编译执行

 

[amao@amao991 mpi-study]$ mpicc scatterReducer.c
[amao@amao991 mpi-study]$ mpiexec -n 4 -f machinefile ./a.out
my rankID = 0, receive Results: 1.000000 2.000000 3.000000 4.000000, totalOnMe = 10.000000
my rankID = 1, receive Results: 5.000000 6.000000 7.000000 8.000000, totalOnMe = 26.000000
Process 1 is done.
my rankID = 3, receive Results: 13.000000 14.000000 15.000000 16.000000, totalOnMe = 58.000000
Process 3 is done.
Process 0 is done.
my rankID = 2, receive Results: 9.000000 10.000000 11.000000 12.000000, totalOnMe = 42.000000
Process 2 is done.
totalSum = 136.000000

 

5. 总结

(1)本例先通过MPI_Scatter函数把数组分发给4个进程(每个进程收到4个float值)，每个进程对自己收到的4个数据求和

(2)然后通过MPI_Reduce函数汇总4个进程的数据，得到总和totalSum值。

(3)本例典型地说明了数据分发个多个进程／每个进程单独求解／然后汇总。

----------------------------------------------------------------

1. MPI_Gather函数

MPI_Gather (&sendbuf,sendcnt,sendtype,&recvbuf, recvcount,recvtype,root,comm) 

Gathers distinct messages from each task in the group to a single destination task. This routine is the reverse operation of MPI_Scatter. 

2. 图示

3. 举例

 

#include<stdio.h>
#include<stdlib.h>
#include"mpi.h"

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

        MPI_Init(&argc, &argv);
        MPI_Barrier(MPI_COMM_WORLD);
        double elapsed_time = -MPI_Wtime();

        MPI_Comm_rank(MPI_COMM_WORLD, &rankID);
        MPI_Comm_size(MPI_COMM_WORLD, &totalNumTasks);

        int* gatherBuf = (int *)malloc(sizeof(int) * totalNumTasks);
        if(gatherBuf == NULL){
                printf("malloc error!");
                exit(-1);
                MPI_Finalize();
        }

        int sendBuf = rankID; //for each process, its rankID will be sent out

        int sendCount = 1;
        int recvCount = 1;
        int root = 0;
        MPI_Gather(&sendBuf, sendCount, MPI_INT, gatherBuf, recvCount, MPI_INT, root, MPI_COMM_WORLD);

        elapsed_time += MPI_Wtime();
        if(!rankID){
                int i;
                for(i = 0; i < totalNumTasks; i++){
                        printf("gatherBuf[%d] = %d, ", i, gatherBuf[i]);
                }
                putchar('\n');
                printf("total elapsed time = %10.6f\n", elapsed_time);
        }

        MPI_Finalize();
        return 0;
}

 

4. 编译执行

 

[amao@amao991 mpi-study]$ mpicc gather.c
[amao@amao991 mpi-study]$ mpiexec -f machinefile -n 5 ./a.out
gatherBuf[0] = 0, gatherBuf[1] = 1, gatherBuf[2] = 2, gatherBuf[3] = 3, gatherBuf[4] = 4,
total elapsed time =   0.009653

 

5. 总结

(1) 本例用了malloc在heap中开辟连续空间(数组)

int* gatherBuf = (int *)malloc(sizeof(int) * totalNumTasks);     而后用gatherBuf[i]的形式来逐个显示每个空间的值。