到目前为止,我有以下代码。这将生成笛卡尔通信器,找出MPI进程的坐标并计算出基于此应该获得多少数组(因为数组不需要是笛卡尔网格大小的倍数)。然后,我创建了一个新的MPI派生数据类型,它将整个进程的子数组作为一个项目发送(也就是说,每个进程的步幅、块长度和计数都不同,因为每个进程具有不同大小的数组)。但是,当我使用MPI_Gather将数据聚集在一起时,我会遇到分段错误。
我认为这是因为我不应该在MPI_Gather调用中使用相同的数据类型进行发送和接收。数据类型对于发送数据来说是正确的,因为它具有正确的计数、步幅和块长度,但是当它到达另一端时,它将需要非常不同的派生数据类型。我不确定如何计算此数据类型的参数 - 有人有任何想法吗?
此外,如果我完全从错误的角度来处理此问题,请告诉我!
#include<stdio.h>
#include<array_alloc.h>
#include<math.h>
#include<mpi.h>
int main(int argc, char ** argv)
{
int size, rank;
int dim_size[2];
int periods[2];
int A = 2;
int B = 3;
MPI_Comm cart_comm;
MPI_Datatype block_type;
int coords[2];
float **array;
float **whole_array;
int n = 10;
int rows_per_core;
int cols_per_core;
int i, j;
int x_start, x_finish;
int y_start, y_finish;
/* Initialise MPI */
MPI_Init(&argc, &argv);
/* Get the rank for this process, and the number of processes */
MPI_Comm_size(MPI_COMM_WORLD, &size);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
if (rank == 0)
{
/* If we're the master process */
whole_array = alloc_2d_float(n, n);
/* Initialise whole array to silly values */
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
whole_array[i][j] = 9999.99;
}
}
for (j = 0; j < n; j ++)
{
for (i = 0; i < n; i++)
{
printf("%f ", whole_array[j][i]);
}
printf("\n");
}
}
/* Create the cartesian communicator */
dim_size[0] = B;
dim_size[1] = A;
periods[0] = 1;
periods[1] = 1;
MPI_Cart_create(MPI_COMM_WORLD, 2, dim_size, periods, 1, &cart_comm);
/* Get our co-ordinates within that communicator */
MPI_Cart_coords(cart_comm, rank, 2, coords);
rows_per_core = ceil(n / (float) A);
cols_per_core = ceil(n / (float) B);
if (coords[0] == (B - 1))
{
/* We're at the far end of a row */
cols_per_core = n - (cols_per_core * (B - 1));
}
if (coords[1] == (A - 1))
{
/* We're at the bottom of a col */
rows_per_core = n - (rows_per_core * (A - 1));
}
printf("X: %d, Y: %d, RpC: %d, CpC: %d\n", coords[0], coords[1], rows_per_core, cols_per_core);
MPI_Type_vector(rows_per_core, cols_per_core, cols_per_core + 1, MPI_FLOAT, &block_type);
MPI_Type_commit(&block_type);
array = alloc_2d_float(rows_per_core, cols_per_core);
if (array == NULL)
{
printf("Problem with array allocation.\nExiting\n");
return 1;
}
for (j = 0; j < rows_per_core; j++)
{
for (i = 0; i < cols_per_core; i++)
{
array[j][i] = (float) (i + 1);
}
}
MPI_Barrier(MPI_COMM_WORLD);
MPI_Gather(array, 1, block_type, whole_array, 1, block_type, 0, MPI_COMM_WORLD);
/*
if (rank == 0)
{
for (j = 0; j < n; j ++)
{
for (i = 0; i < n; i++)
{
printf("%f ", whole_array[j][i]);
}
printf("\n");
}
}
*/
/* Close down the MPI environment */
MPI_Finalize();
}
我使用的2D数组分配程序实现如下:
float **alloc_2d_float( int ndim1, int ndim2 ) {
float **array2 = malloc( ndim1 * sizeof( float * ) );
int i;
if( array2 != NULL ){
array2[0] = malloc( ndim1 * ndim2 * sizeof( float ) );
if( array2[ 0 ] != NULL ) {
for( i = 1; i < ndim1; i++ )
array2[i] = array2[0] + i * ndim2;
}
else {
free( array2 );
array2 = NULL;
}
}
return array2;
}
alloc_2d_float
的实现吗? - suszterpatt