我是mpi编程的新手。我试图写矩阵乘法。通过使用分散和收集例程的关于矩阵乘法的帖子MPI Matrix Multiplication with scatter gather。 我尝试修改上面帖子中提供的代码,如下所示......
#define N 4
#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stddef.h>
#include "mpi.h"
void print_results(char *prompt, int a[N][N]);
int main(int argc, char *argv[])
{
int i, j, k, rank, size, tag = 99, blksz, sum = 0;
int a[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
int b[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
int c[N][N];
int aa[N],cc[N];
MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &size);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
//scatter rows of first matrix to different processes
MPI_Scatter(a, N*N/size, MPI_INT, aa, N*N/size, MPI_INT,0,MPI_COMM_WORLD);
//broadcast second matrix to all processes
MPI_Bcast(b, N*N, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Barrier(MPI_COMM_WORLD);
//perform vector multiplication by all processes
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
{
sum = sum + aa[j] * b[i][j];
}
cc[i] = sum;
sum = 0;
}
MPI_Gather(cc, N*N/size, MPI_INT, c, N*N/size, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Barrier(MPI_COMM_WORLD);
MPI_Finalize();
print_results("C = ", c);
}
void print_results(char *prompt, int a[N][N])
{
int i, j;
printf ("\n\n%s\n", prompt);
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
printf(" %d", a[i][j]);
}
printf ("\n");
}
printf ("\n\n");
}
我在
上面运行了程序$mpirun -np 4 ./a.out
对于上述程序,我收到了错误的输出..
C =
0 0 -562242168 32766
1 0 4197933 0
-562242176 32766 0 0
4197856 0 4196672 0
C =
0 0 -1064802792 32765
1 0 4197933 0
-1064802800 32765 0 0
4197856 0 4196672 0
C =
30 70 29 60
70 174 89 148
29 89 95 74
60 148 74 126
C =
0 0 -1845552920 32765
1 0 4197933 0
-1845552928 32765 0 0
4197856 0 4196672 0
我有以下查询 1.为什么结果矩阵C被所有进程打印。它是 应该只由主要过程打印。 2.为什么打印出错误的结果?
对此方面的更正和帮助表示赞赏。
答案 0 :(得分:5)
结果矩阵c
由所有进程打印,因为每个进程都执行函数void print_results(char *prompt, int a[N][N])
。由于您正在收集排名为0的进程,因此在调用if (rank == 0)
函数之前添加语句print_results(...)
。此外,由于:
for (j = 0; j < N; j++)
{
sum = sum + aa[j] * b[i][j];
}
这应该是:
for (j = 0; j < N; j++)
{
sum = sum + aa[j] * b[j][i];
}
此外,无需广播b
,因为所有流程都已有一份副本,您可以避免MPI_Barrier()
。完整的程序然后变成:
#define N 4
#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stddef.h>
#include "mpi.h"
void print_results(char *prompt, int a[N][N]);
int main(int argc, char *argv[])
{
int i, j, k, rank, size, tag = 99, blksz, sum = 0;
int a[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
int b[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
int c[N][N];
int aa[N],cc[N];
MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &size);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
//scatter rows of first matrix to different processes
MPI_Scatter(a, N*N/size, MPI_INT, aa, N*N/size, MPI_INT,0,MPI_COMM_WORLD);
//broadcast second matrix to all processes
MPI_Bcast(b, N*N, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Barrier(MPI_COMM_WORLD);
//perform vector multiplication by all processes
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
{
sum = sum + aa[j] * b[j][i]; //MISTAKE_WAS_HERE
}
cc[i] = sum;
sum = 0;
}
MPI_Gather(cc, N*N/size, MPI_INT, c, N*N/size, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Barrier(MPI_COMM_WORLD);
MPI_Finalize();
if (rank == 0) //I_ADDED_THIS
print_results("C = ", c);
}
void print_results(char *prompt, int a[N][N])
{
int i, j;
printf ("\n\n%s\n", prompt);
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
printf(" %d", a[i][j]);
}
printf ("\n");
}
printf ("\n\n");
}
然后c =
C =
54 37 47 57
130 93 119 145
44 41 56 71
111 79 101 123
答案 1 :(得分:0)
调用mpi_finalize并不表示所有MPI进程都像在OpenMP中那样终止!
在大多数mpi实现中,所有进程都在MPI_init之前和MPI_Finalized之后执行指令。
一个好的实践是在MPI_Init之前和MPI_Finalized之后不执行任何操作。