我尝试将一个矩阵逐行散布到所有处理器但是它导致了分段错误。我不知道我做错了什么..这是我的代码
if(rank == 0) {
A_row = 10;
A_col = 10;
/* calculate the strip size */
strip_size = A_row / size;
/* genarate Matrix A */
A = (double **)malloc(sizeof(double*) * 10);
int k = 0;
for(i = 0; i < 10; i++) {
A[i] = (double*)malloc(sizeof(double) * 10);
for(j = 0; j < 10; j++) {
A[i][j] = k;
k++;
printf("%lf ", A[i][j]);
}
printf("\n");
}
}
/* Broadcasting the row, column size of Matrix A as well as strip size and Matrix B*/
MPI_Bcast(&A_row, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&A_col, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&strip_size, 1, MPI_INT, 0, MPI_COMM_WORLD);
/* defining a datatype for sub-matrix */
MPI_Type_vector(strip_size, A_col, A_col, MPI_DOUBLE, &strip);
MPI_Type_commit(&strip);
strip_A = (double **)malloc(sizeof(double*)*strip_size);
for(i= 0; i< strip_size; i++) {
strip_A[i] = (double*)malloc(sizeof(double)*A_col);
}
MPI_Scatter(&A[0][0], 1, strip, &strip_A[0][0], 1, strip, 0, MPI_COMM_WORLD);
for(i = 0; i < strip_size; i++) {
if(i == 0) {
printf("rank = %d\n", rank);
}
for(j = 0; j < A_col; j++) {
printf("%lf ", strip_A[i][j]);
}
printf("\n");
}
谁能告诉我有什么问题......
这是我运行时的错误
mpirun -np 2 ./a.out
0.000000 1.000000 2.000000 3.000000 4.000000 5.000000 6.000000 7.000000 8.000000 9.000000
10.000000 11.000000 12.000000 13.000000 14.000000 15.000000 16.000000 17.000000 18.000000 19.000000
20.000000 21.000000 22.000000 23.000000 24.000000 25.000000 26.000000 27.000000 28.000000 29.000000
30.000000 31.000000 32.000000 33.000000 34.000000 35.000000 36.000000 37.000000 38.000000 39.000000
40.000000 41.000000 42.000000 43.000000 44.000000 45.000000 46.000000 47.000000 48.000000 49.000000
50.000000 51.000000 52.000000 53.000000 54.000000 55.000000 56.000000 57.000000 58.000000 59.000000
60.000000 61.000000 62.000000 63.000000 64.000000 65.000000 66.000000 67.000000 68.000000 69.000000
70.000000 71.000000 72.000000 73.000000 74.000000 75.000000 76.000000 77.000000 78.000000 79.000000
80.000000 81.000000 82.000000 83.000000 84.000000 85.000000 86.000000 87.000000 88.000000 89.000000
90.000000 91.000000 92.000000 93.000000 94.000000 95.000000 96.000000 97.000000 98.000000 99.000000
rank = 1
42.000000 43.000000 44.000000 45.000000 46.000000 47.000000 48.000000 49.000000 0.000000 0.000000
52.000000 53.000000 54.000000 55.000000 56.000000 57.000000 58.000000 59.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
[seismicmstm:07338] *** Process received signal *** [seismicmstm:07338] Signal:
Segmentation fault (11)
[seismicmstm:07338] Signal code:
(128) [seismicmstm:07338] Failing at
address: (nil)
-------------------------------------------------------------------------- mpirun noticed that process rank 1 with PID 7338 on node seismicmstm.cluster exited on signal 11 (Segmentation fault).
--------------------------------------------------------------------------
答案 0 :(得分：6)
这里有几件事情要发生。好消息是，最难的东西 - 创建mpi数据类型，以及MPI_Scatter调用的基本结构 - 都是正确的。
第一个问题是MPI_Scatter行使用＆amp;（A [0] [0]） - 但除了排名为零之外，你还没有设置A指向任何东西！所以你取消引用一个随机指针两次，这就是你的段错误。
正如suszterpatt所建议的那样，一个更微妙的问题是，无法保证您的已分配内存行是连续的，因此即使您修复了上述操作，分散操作也可能无法正常工作。你试图将strip_size * A_col从A中的某个地方发送到strip_A，但是strip_A可能不是连续组成的那么多双 - 它可能是A_col加倍，然后是一些填充，然后A_col加倍 - 或者实际上，各行可以分散在各处。固定的三种方法是，为了方便起见（IMHO）:( a）通过创建整个数组然后创建指向正确位置的二维C数组，使数据在内存中连续; （b）一次只发一排;或者（c）创建一个MPI数据类型，实际上反映了数据在内存中的映射方式（可能是随机的）。
使用（a）似乎有效的方法（A_row按大小均分），如下所示：
#include <stdio.h>
#include <mpi.h>
#include <stdlib.h>
int main(int argc, char** argv) {
int rank, size;
int strip_size, A_row, A_col;
double **A, **strip_A, *Adata, *stripdata;
MPI_Datatype strip;
int i,j;
MPI_Init(&argc,&argv) ;
MPI_Comm_rank(MPI_COMM_WORLD,&rank) ;
MPI_Comm_size(MPI_COMM_WORLD,&size) ;
if(rank == 0) {
A_row = 10;
A_col = 10;
/* calculate the strip size */
strip_size = A_row / size;
/* genarate Matrix A */
Adata = (double *)malloc(sizeof(double)*A_row*A_col);
A = (double **)malloc(sizeof(double*) * A_row);
for(i = 0; i < A_row; i++) {
A[i] = &(Adata[i*A_col]);
}
int k = 0;
for(i = 0; i < A_row; i++) {
for(j = 0; j < A_col; j++) {
A[i][j] = k;
k++;
}
}
}
/* Broadcasting the row, column size of Matrix A as well as strip size and Matrix B*/
MPI_Bcast(&A_row, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&A_col, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&strip_size, 1, MPI_INT, 0, MPI_COMM_WORLD);
/* defining a datatype for sub-matrix */
MPI_Type_vector(strip_size, A_col, A_col, MPI_DOUBLE, &strip);
MPI_Type_commit(&strip);
stripdata = (double *)malloc(sizeof(double)*strip_size*A_col);
strip_A = (double **)malloc(sizeof(double*)*strip_size);
for(i= 0; i< strip_size; i++) {
strip_A[i] = &(stripdata[i*A_col]);
}
MPI_Scatter(Adata, 1, strip, &(strip_A[0][0]), 1, strip, 0, MPI_COMM_WORLD);
//MPI_Scatter(Adata, A_col*strip_size, MPI_DOUBLE, &(strip_A[0][0]), A_col*strip_size, MPI_DOUBLE, 0, MPI_COMM_WORLD);
for(i = 0; i < strip_size; i++) {
if(i == 0) {
printf("rank = %d\n", rank);
}
for(j = 0; j < A_col; j++) {
printf("%lf ", strip_A[i][j]);
}
printf("\n");
}
MPI_Type_free(&strip);
free(strip_A);
free(stripdata);
free(Adata);
free(A);
return 0;
}
答案 1 :(得分：1)
我认为最终，你所做错的是将矩阵存储为数组数组。我想你会发现，如果你把它存储在一个单独的数组中（按行主要或列主要顺序，哪个适合你的想象），事情会变得容易多了。
答案 2 :(得分：0)
只需添加MPI_Finalize（）;对你的命令。 ;） 请参阅下面的代码并输出。输出正确但由于屏障而无法正确打印。要么你可以使用MPI_Barrier（）或使用MPI_Isend（）和MPI_Irecv（）。享受
#include <stdio.h>
#include <mpi.h>
#include <stdlib.h>
int main(int argc, char** argv) {
int rank, size;
int strip_size, A_row, A_col;
double **A, **strip_A, *Adata, *stripdata;
MPI_Datatype strip;
int i,j;
MPI_Init(&argc,&argv) ;
MPI_Comm_rank(MPI_COMM_WORLD,&rank) ;
MPI_Comm_size(MPI_COMM_WORLD,&size) ;
if(rank == 0) {
A_row = 10;
A_col = 10;
/* calculate the strip size */
strip_size = A_row / size;
/* genarate Matrix A */
Adata = (double *)malloc(sizeof(double)*A_row*A_col);
A = (double **)malloc(sizeof(double*) * A_row);
for(i = 0; i < A_row; i++) {
A[i] = &(Adata[i*A_col]);
}
int k = 0;
for(i = 0; i < A_row; i++) {
for(j = 0; j < A_col; j++) {
A[i][j] = k;
k++;
}
}
}
/* Broadcasting the row, column size of Matrix A as well as strip size and Matrix B*/
MPI_Bcast(&A_row, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&A_col, 1, MPI_INT, 0, MPI_COMM_WORLD);
MPI_Bcast(&strip_size, 1, MPI_INT, 0, MPI_COMM_WORLD);
/* defining a datatype for sub-matrix */
MPI_Type_vector(strip_size, A_col, A_col, MPI_DOUBLE, &strip);
MPI_Type_commit(&strip);
stripdata = (double *)malloc(sizeof(double)*strip_size*A_col);
strip_A = (double **)malloc(sizeof(double*)*strip_size);
for(i= 0; i< strip_size; i++) {
strip_A[i] = &(stripdata[i*A_col]);
}
MPI_Scatter(Adata, 1, strip, &(strip_A[0][0]), 1, strip, 0, MPI_COMM_WORLD);
//MPI_Scatter(Adata, A_col*strip_size, MPI_DOUBLE, &(strip_A[0][0]), A_col*strip_size, MPI_DOUBLE, 0, MPI_COMM_WORLD);
for(i = 0; i < strip_size; i++) {
if(i == 0) {
printf("rank = %d\n", rank);
}
for(j = 0; j < A_col; j++) {
printf("%lf ", strip_A[i][j]);
}
printf("\n");
}
if(rank == 0){
MPI_Type_free(&strip);
free(strip_A);
free(stripdata);
free(Adata);
free(A);}
MPI_Finalize();
return 0;
}
输出
rank = 0
0.000000 1.000000 2.000000 3.000000 4.000000 5.000000 6.000000 7.000000 8.000000 9.000000
rank = 2
20.000000 21.000000 22.000000 23.000000 24.000000 25.000000 26.000000 27.000000 28.000000 29.000000
rank = 6
60.000000 61.000000 62.000000 63.000000 64.000000 65.000000 66.000000 67.000000 68.000000 69.000000
rank = 1
10.000000 11.000000 12.000000 13.000000 14.000000 15.000000 16.000000 17.000000 18.000000 19.000000
rank = 3
30.000000 31.000000 32.000000 33.000000 34.000000 35.000000 36.000000 37.000000 38.000000 39.000000
rank = 5
50.000000 51.000000 52.000000 53.000000 54.000000 55.000000 56.000000 57.000000 58.000000 59.000000
rank = 8
80.000000 81.000000 82.000000 83.000000 84.000000 85.000000 86.000000 87.000000 88.000000 89.000000
rank = 7
70.000000 71.000000 72.000000 73.000000 74.000000 75.000000 76.000000 77.000000 78.000000 79.000000
rank = 9
90.000000 91.000000 92.000000 93.000000 94.000000 95.000000 96.000000 97.000000 98.000000 99.000000
rank = 4
40.000000 41.000000 42.000000 43.000000 44.000000 45.000000 46.000000 47.000000 48.000000 49.000000