ok so I am trying to send a structure like so over MPI
struct BColumns {
double **B;
int offset;
};
And if I just do some BS allocation of data like so
bSet.offset = myRank;
bSet.B = (double **) calloc(2, sizeof(double *));
bSet.B[0] = (double *) calloc(1, sizeof(double));
bSet.B[1] = (double *) calloc(1, sizeof(double));
bSet.B[0][0] = 1;
bSet.B[1][0] = 2;
if(myRank == 0){
MPI_Send(&bSet,sizeof(struct BColumns), MPI_BYTE, 1, 1, MPI_COMM_WORLD);
}else{
MPI_Recv(&recvBuf, sizeof(struct BColumns), MPI_BYTE, MPI_ANY_SOURCE, MPI_ANY_TAG, MPI_COMM_WORLD, &status );
}
And I am assuming that its not going to work well because if I send this structure as is it will just send the pointer in B and that pointer doesn't point to anything on the other processor, so how would I go about sending data like this in MPI.
As suszterpatt points out, you really want to allocate your block of B in one big chunk; that's probably better for performance anyway but it's really required for any communications so you're not chasing pointers everywhere. And I think one way or another you're probably going to have to do it in different sends -- sending size information, then the data in one chunk -- although you could probably create and delete a different MPI_Type_struct for every one of these you send. But using multiple sends per object isn't very hard:
#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
typedef struct BColumns {
double **B;
int offset;
} bc;
double **alloc2d(int n, int m) {
double *data = malloc(n*m*sizeof(double));
double **array = malloc(n*sizeof(double *));
for (int i=0; i<n; i++) {
array[i] = &(data[i*m]);
}
return array;
}
void free2d(double **array) {
free(array[0]);
free(array);
}
int main(int argc, char **argv) {
const int tag = 13;
int size, rank;
MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &size);
if (size < 2) {
fprintf(stderr,"Requires at least two processes.\n");
exit(-1);
}
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
if (rank == 0) {
int ncols=3, colsize=5;
bc *send;
send = malloc(sizeof(bc));
send->offset = 1;
send->B = alloc2d(ncols, colsize);
for (int i=0; i<ncols; i++)
for (int j=0; j<colsize; j++)
send->B[i][j] = i*j;
const int dest = 1;
MPI_Send(&ncols, 1, MPI_INT, dest, tag, MPI_COMM_WORLD);
MPI_Send(&colsize, 1, MPI_INT, dest, tag, MPI_COMM_WORLD);
MPI_Send(&(send->offset), 1, MPI_INT, dest, tag, MPI_COMM_WORLD);
MPI_Send(&(send->B[0][0]), ncols*colsize, MPI_DOUBLE, dest, tag,
MPI_COMM_WORLD);
printf("Rank %d: sent structure B\n", rank);
free2d(send->B);
free(send);
}
if (rank == 1) {
MPI_Status status;
const int src=0;
int rncols, rcolsize;
bc *recv;
MPI_Recv(&rncols, 1, MPI_INT, src, tag, MPI_COMM_WORLD, &status);
MPI_Recv(&rcolsize, 1, MPI_INT, src, tag, MPI_COMM_WORLD, &status);
printf("Rank %d: Received: rncols = %d rcolsize=%d\n", rank, rncols, rcolsize);
recv = malloc(sizeof(bc));
recv->B = alloc2d(rncols, rcolsize);
MPI_Recv(&(recv->offset), 1, MPI_INT, src, tag, MPI_COMM_WORLD, &status);
MPI_Recv(&(recv->B[0][0]), rncols*rcolsize, MPI_DOUBLE, src, tag,
MPI_COMM_WORLD, &status);
printf("Rank %d: Received: offset = %d\n", rank, recv->offset);
for (int i=0; i<rncols; i++) {
printf("%d: Column %d/%d: ", rank, i, rncols);
for (int j=0; j<rcolsize; j++)
printf(" %lf ", recv->B[i][j]);
printf("\n");
}
free2d(recv->B);
free(recv);
}
MPI_Finalize();
return 0;
}
And then running it:
$ mpirun -np 3 ./bstruct
Rank 0: sent structure B
Rank 1: Received: rncols = 3 rcolsize=5
Rank 1: Received: offset = 1
1: Column 0/3: 0.000000 0.000000 0.000000 0.000000 0.000000
1: Column 1/3: 0.000000 1.000000 2.000000 3.000000 4.000000
1: Column 2/3: 0.000000 2.000000 4.000000 6.000000 8.000000
You could marshall that data into one message if you wanted to avoid the latency of several sends (and if you knew before hand a maximum size for the B array) either by hand or using MPI function calls or data types, but you'd still have to do it in a similar way.