我正在尝试将多个“B”矩阵列发送到处理器0的不同处理器。我正在尝试发送使用MPI_Send但它不起作用。有人可以帮助我吗?
例如:方阵B的大小为7。 这样就应该分发。
处理器0:3列
处理器1:2列
处理器2:2列
#include <stdlib.h>
#include <mpi.h>
#include <stdio.h>
#define ERR_BADORDER 255
#define TAG_INIT 31337
#define TAG_RESULT 42
#define DISP_MAXORDER 12
int mm(double *A, double *B, double *C, int n, int n1);
int rc(int rt,int rank, int size);
int main(int argc, char *argv[]) {
double *A, *B, *C,t,tt;
int n = 0, n0, n1, n2, i,ss,sts;
int rank = 0, size = 1,prev,next,k,z,jcol,ix=0,m,j;
MPI_Datatype column;
MPI_Request reqs[4];
MPI_Status stats[2];
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &size);
if (!rank) {
if (argc > 1) {
n = atoi(argv[1]);
}
}
MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);
if (!n) {
MPI_Finalize();
return 0;
}
n1 = rc(n, rank,size);
n0 = n * n1;
n2 = n * n;
A = (double *) malloc(sizeof(double) * (rank ? n0 : n2));
B = (double *) malloc(sizeof(double) * n2 );
C = (double *) malloc(sizeof(double) * (rank ? n0 : n2));
if (!rank) {
for (i=0; i<n2; i++) {
A[i] = 1.0;
B[i] = 1.0;
}
}
t = MPI_Wtime();
if (!rank) {
ss = n0;
for (i=1; i<size; i++) {
sts = n * rc(n, i, size);
MPI_Send(A + ss, sts, MPI_DOUBLE, i, TAG_INIT,
MPI_COMM_WORLD);
ss += sts;
}
}
else {
MPI_Recv(A, n0, MPI_DOUBLE, 0, TAG_INIT, MPI_COMM_WORLD,
MPI_STATUS_IGNORE);
}
MPI_Type_vector(n,1,n,MPI_DOUBLE, &column);
MPI_Type_commit(&column);
if (!rank) {
for (i=1; i<size; i++) {
for(m=0;m<=i-1;m++)
ix+=rc(n,m,size);
ss=rc(n,i,size);
for(j=ix;j<ss+ix;j++)
MPI_Send(&B[j], 1, column, i, TAG_INIT, MPI_COMM_WORLD);
/* MPI_Send(&B[i+(n-1)*n], 1, column, i, TAG_INIT,
MPI_COMM_WORLD);*/
}
}
else {
printf("hello");
MPI_Recv(B, n, MPI_DOUBLE, 0, TAG_INIT, MPI_COMM_WORLD,
MPI_STATUS_IGNORE);
}
for (i=0; i<n0; i++) {
printf("Processor: %d and matrix %lf \n ",rank, B[i]);
}
for (i=0; i<n0; i++) {
C[i] = 0.0;
}
MPI_Finalize();
return 0;
}
int rc(int rt, int rank, int size) {
return (rt / size) + (rt % size > rank);
}
答案 0 :(得分:0)
请不要用两三个字母来调用这些值,因为我无法理解你想做什么。 您可以通过不同方式解决问题。 当n = 7且我有3个进程时,我向每个进程发送2列不同于主进程0.我希望这对你有所帮助。最好的问候。
#include <stdlib.h>
#include <mpi.h>
#include <stdio.h>
#define ERR_BADORDER 255
#define TAG_INIT 31337
#define TAG_RESULT 42
#define DISP_MAXORDER 12
int main(int argc, char *argv[]) {
double *B;
int n = 0;
int rank , size;
int i;
int columnToSend;
MPI_Datatype column;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &size);
if (!rank)printf ("rank = %d , size = %d\n", rank, size);
if (!rank) {
if (argc > 1) {
n = atoi(argv[1]);
}
}
MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);
if (!n) {
printf ("n = %d!!\n", n);
MPI_Finalize();
return 0;
}
int offset = n%size;
int gap = n/size;
if (!rank)printf ("n = %d, offset = %d , gap = %d\n", n, offset, gap);
MPI_Type_vector(n,gap,n,MPI_DOUBLE, &column);
MPI_Type_commit(&column);
B = (double *) malloc(sizeof(double) * n*n );
for (i = 0 ; i < n * n ; i++) {
B[i] = -1.0;
}
if (!rank) {
for (i = 0 ; i < n * n ; i++) {
B[i] = i;//<----- I put i instead one
}
for (i=1; i < size; i++) {
columnToSend = gap *i + offset;
printf ("columnToSend = %d to i = %d \n", columnToSend, i);
MPI_Send(&B[columnToSend], 1, column, i, TAG_INIT, MPI_COMM_WORLD);
}
}
if (rank) {
printf ("in rank = %d n*gap = %d \n", rank, n*gap);
MPI_Recv(B, n*gap, MPI_DOUBLE, 0, TAG_INIT, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
for (i=0; i < n*gap; i++) {
printf("Processor: %d and matrix %lf \n ",rank, B[i]);
}
}
MPI_Finalize();
return 0;
}