MPI中的动态内存分配

Question

我是MPI的新手.我写了一个简单的代码来显示使用多个进程的矩阵.如果我有一个8x8的矩阵并用4个进程启动MPI程序,那么我的第1个进程将打印第2个2行,第2个线程将由第2个线程打印,然后将它自己分开.

#define S 8

MPI_Status status;

int main(int argc, char *argv[])
{
int numtasks, taskid;
int i, j, k = 0;

MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &taskid);
MPI_Comm_size(MPI_COMM_WORLD, &numtasks);

int rows, offset, remainPart, orginalRows, height, width;
int **a;
//  int a[S][S];

if(taskid == 0)
{
    cout<<taskid<<endl;
    height = width = S;

    a = (int **)malloc(height*sizeof(int *));
    for(i=0; i<height; i++)
        a[i] =  (int *)malloc(width*sizeof(int));

    for(i=0; i<S; i++)
        for(j=0; j<S; j++)
            a[i][j] = ++k;

    rows = S/numtasks;
    offset = rows;
    remainPart = S%numtasks;

    cout<<"Num Rows : "<<rows<<endl;

    for(i=1; i<numtasks; i++)
        if(remainPart > 0)
        {
            orginalRows = rows;
            rows++;
            remainPart--;

            MPI_Send(&offset, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&rows, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&width, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&a[offset][0], rows*S, MPI_INT,i,1, MPI_COMM_WORLD);

            offset += rows;
            rows = orginalRows;
        }
        else
        {
            MPI_Send(&offset, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&rows, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&width, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
            MPI_Send(&a[offset][0], rows*S, MPI_INT,i,1, MPI_COMM_WORLD);

            offset += rows;
        }

        //Processing
        rows = S/numtasks;
        for(i=0; i<rows; i++)
        {
            for(j=0; j<width; j++)
                cout<<a[i][j]<<"\t";
            cout<<endl;
        }
}else
{
    cout<<taskid<<endl;

    MPI_Recv(&offset, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
    MPI_Recv(&rows, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
    MPI_Recv(&width, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
    a = (int **)malloc(rows*sizeof(int *));
    for(i=0; i<rows; i++)
        a[i] =  (int *)malloc(width*sizeof(int));
    MPI_Recv(&a, rows*width, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
    cout<<"Offset : "<<offset<<"\nRows : "<<rows<<"\nWidth : "<<width<<endl;

    for(i=0; i<rows; i++)
    {
        for(j=0; j<width; j++)
            cout<<a[i][j]<<"\t";
        cout<<endl;
    }
}

getch();
MPI_Finalize();

return 0;
}

这是我的完整代码,这里我已经为'a'动态分配了内存,而在打印[i] [j]时,在else部分下,我收到了运行时错误.如果我将动态内存分配更改为静态,如将int**a更改为int a [N] [N]并删除

    a = (int **)malloc(rows*sizeof(int));
    for(i=0; i<rows; i++)
        a[i] =  (int *)malloc(width*sizeof(int));

它完美地运作.

Answer 1

至少有两种方法可以动态分配2D阵列.

第一个是@HRoid:每行一次分配一个.看看这里获得一个计划.

第二个是@Claris建议的,它将确保数据在内存中是连续的.这是许多MPI操作所要求的......如FFTW(2D快速傅立叶变换)或Lapack(线性代数的密集矩阵)等库也需要它.您的程序可能会失败

MPI_Send(&a[offset][0], rows*S, MPI_INT,i,1, MPI_COMM_WORLD);

如果S>1,此程序将尝试发送行结束后的项目n° offset...这可能会触发分段错误或未定义的行为.

您可以分配您的阵列这样:

a = malloc(rows * sizeof(int *));
if(a==NULL){fprintf(stderr,"out of memory...i will fail\n");}
int *t = malloc(rows * width * sizeof(int));
if(t==NULL){fprintf(stderr,"out of memory...i will fail\n");}
for(i = 0; i < rows; ++i)
  a[i] = &t[i * width];

注意:malloc 不将内存初始化为0!

您似乎希望在许多过程中传播2D数组.看看MPI_Scatterv() 这里.看看这个问题.

如果您想了解有关2D阵列和MPI的更多信息,请查看此处.

您可以在此处找到MPI_Scatterv的基本示例.

我改变#define S 8了#define SQUARE_SIZE 42.提供描述性名称总是更好.

这是一个使用MPI_Scatterv()的工作代码!

#include <mpi.h>
#include <iostream> 
#include <cstdlib>

using namespace std;

#define SQUARE_SIZE 42

MPI_Status status;

int main(int argc, char *argv[])
{
    int numtasks, taskid;
    int i, j, k = 0;

    MPI_Init(&argc, &argv);
    MPI_Comm_rank(MPI_COMM_WORLD, &taskid);
    MPI_Comm_size(MPI_COMM_WORLD, &numtasks);

    int rows, offset, remainPart, orginalRows, height, width;
    int **a;

    height = width = SQUARE_SIZE;

    //on rank 0, let's build a big mat of int
    if(taskid == 0){ 
        a=new int*[height]; 
        int *t =new int[height * width];
        for(i = 0; i < height; ++i)
            a[i] = &t[i * width];
        for(i=0; i<height; i++)
            for(j=0; j<width; j++)
                a[i][j] = ++k;
    }

    //for everyone, lets compute numbers of rows, numbers of int and displacements for everyone. Only 0 will use these arrays, but it's a practical way to get `rows` 
    int nbrows[numtasks];
    int sendcounts[numtasks];
    int displs[numtasks];
    displs[0]=0;
    for(i=0;i<numtasks;i++){
        nbrows[i]=height/numtasks;
        if(i<height%numtasks){
            nbrows[i]=nbrows[i]+1;
        }
        sendcounts[i]=nbrows[i]*width;
        if(i>0){
            displs[i]=displs[i-1]+sendcounts[i-1];
        }
    }
    rows=nbrows[taskid];

    //scattering operation. 
    //The case of the root is particular, since the communication is not to be done...Hence, the flag MPI_IN_PLACE is used.
    if(taskid==0){
        MPI_Scatterv(&a[0][0],sendcounts,displs,MPI_INT,MPI_IN_PLACE,0,MPI_INT,0,MPI_COMM_WORLD);
    }else{
        //allocation of memory for the piece of mat on the other nodes.
        a=new int*[rows];
        int *t =new int[rows * width];
        for(i = 0; i < rows; ++i)
            a[i] = &t[i * width];

        MPI_Scatterv(NULL,sendcounts,displs,MPI_INT,&a[0][0],rows*width,MPI_INT,0,MPI_COMM_WORLD);
    }
    //printing, one proc at a time
    if(taskid>0){
        MPI_Status status;
        MPI_Recv(NULL,0,MPI_INT,taskid-1,0,MPI_COMM_WORLD,&status);
    }
    cout<<"rank"<< taskid<<" Rows : "<<rows<<" Width : "<<width<<endl;

    for(i=0; i<rows; i++)
    {
        for(j=0; j<width; j++)
            cout<<a[i][j]<<"\t";
        cout<<endl;
    }
    if(taskid<numtasks-1){
        MPI_Send(NULL,0,MPI_INT,taskid+1,0,MPI_COMM_WORLD);
    }

    //freeing the memory !

    delete[] a[0];
    delete[] a;

    MPI_Finalize();

    return 0;
}

编译 : mpiCC main.cpp -o main

跑步 : mpiexec -np 3 main