MPI错误:内存不足 - 有哪些解决方案选项

Question

我正在尝试解决MPI_Irecv中的致命错误:中止作业并收到对该查询的混合(有用但不完整)响应.

该错误消息是以下内容:

aborting job:
> Fatal error in MPI_Irecv: Other MPI
> error, error stack: MPI_Irecv(143):
> MPI_Irecv(buf=0x8294a60, count=48,
> MPI_DOUBLE, src=2, tag=-1, 
> MPI_COMM_WORLD, request=0xffffd6ac)
> failed MPID_Irecv(64): Out of
> memory

我正在寻求某人的帮助来回答这些问题(我需要指导以帮助调试和解决此死锁)

1. 在"MPI非阻塞发送和接收"结束时,发送/接收完成后是自己释放的内存还是必须强制释放它？

2. 如果我使用"多核心"而不是单核心,会解决"内存不足"的问题吗？我们目前有4个处理器到1个核心,我使用以下命令提交我的工作:mpirun -np 4.我尝试使用mpirun n -4但它仍然在同一个核心上运行了4个线程.

3. 我如何计算出我的程序需要多少"共享内存"？

MPI_ISend/MPI_IRecv在我的代码中的递归循环内部,因此如果错误源位于那里则不太清楚(如果我只使用一次或两次发送/接收命令,系统计算就好了没有"内存不足问题"如果是这样,如何检查和减轻此类信息？

 #include <mpi.h>  

 #define Rows 48 

double *A = new double[Rows];
double *AA = new double[Rows];
....
....

 int main (int argc, char *argv[])
 {
  MPI_Status status[8]; 
  MPI_Request request[8];
  MPI_Init (&argc, &argv);
  MPI_Comm_size(MPI_COMM_WORLD, &p);   
  MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);

 while (time < final time){
...
...

for (i=0; i<Columns; i++) 
{
for (y=0; y<Rows; y++) 
{
if ((my_rank) == 0)
{
    MPI_Isend(A, Rows, MPI_DOUBLE, my_rank+1, 0, MPI_COMM_WORLD, &request[1]);
    MPI_Irecv(AA, Rows, MPI_DOUBLE, my_rank+1, MPI_ANY_TAG, MPI_COMM_WORLD, &request[3]);
    MPI_Wait(&request[3], &status[3]);  

    MPI_Isend(B, Rows, MPI_DOUBLE, my_rank+2, 0, MPI_COMM_WORLD, &request[5]);
    MPI_Irecv(BB, Rows, MPI_DOUBLE, my_rank+2, MPI_ANY_TAG, MPI_COMM_WORLD, &request[7]);
    MPI_Wait(&request[7], &status[7]);
 }

if ((my_rank) == 1)
{
    MPI_Irecv(CC, Rows, MPI_DOUBLE, my_rank-1, MPI_ANY_TAG, MPI_COMM_WORLD, &request[1]);
    MPI_Wait(&request[1], &status[1]); 
    MPI_Isend(Cmpi, Rows, MPI_DOUBLE, my_rank-1, 0, MPI_COMM_WORLD, &request[3]);

    MPI_Isend(D, Rows, MPI_DOUBLE, my_rank+2, 0, MPI_COMM_WORLD, &request[6]); 
    MPI_Irecv(DD, Rows, MPI_DOUBLE, my_rank+2, MPI_ANY_TAG, MPI_COMM_WORLD, &request[8]);
    MPI_Wait(&request[8], &status[8]);
}

if ((my_rank) == 2)
{
    MPI_Isend(E, Rows, MPI_DOUBLE, my_rank+1, 0, MPI_COMM_WORLD, &request[2]);
    MPI_Irecv(EE, Rows, MPI_DOUBLE, my_rank+1, MPI_ANY_TAG, MPI_COMM_WORLD, &request[4]);
    MPI_Wait(&request[4], &status[4]);

    MPI_Irecv(FF, Rows, MPI_DOUBLE, my_rank-2, MPI_ANY_TAG, MPI_COMM_WORLD, &request[5]);
    MPI_Wait(&request[5], &status[5]);
    MPI_Isend(Fmpi, Rows, MPI_DOUBLE, my_rank-2, 0, MPI_COMM_WORLD, &request[7]);
}

if ((my_rank) == 3)
{
    MPI_Irecv(GG, Rows, MPI_DOUBLE, my_rank-1, MPI_ANY_TAG, MPI_COMM_WORLD, &request[2]);
    MPI_Wait(&request[2], &status[2]);
        MPI_Isend(G, Rows, MPI_DOUBLE, my_rank-1, 0, MPI_COMM_WORLD, &request[4]);

    MPI_Irecv(HH, Rows, MPI_DOUBLE, my_rank-2, MPI_ANY_TAG, MPI_COMM_WORLD, &request[6]);
        MPI_Wait(&request[6], &status[6]); 
        MPI_Isend(H, Rows, MPI_DOUBLE, my_rank-2, 0, MPI_COMM_WORLD, &request[8]);
 }

谢谢!

Answer 1

你的程序中有内存泄漏; 这个:

MPI_Isend(A, Rows, MPI_DOUBLE, my_rank+1, 0, MPI_COMM_WORLD, &request[1]);
MPI_Irecv(AA, Rows, MPI_DOUBLE, my_rank+1, MPI_ANY_TAG, MPI_COMM_WORLD, &request[3]);
MPI_Wait(&request[3], &status[3])

泄漏与MPI_Isend请求相关的资源.你可以在Rows*Columns每次迭代中调用这个次数,大概是多次迭代; 但你只是在等待其中一个请求.你可能需要MPI_Waitall()为这两个请求做一个.

但除此之外,你的程序非常混乱.没有明智的MPI计划应该有这样一系列的if (rank == ...)陈述.既然你没有在非阻塞发送/接收和等待之间做任何实际的工作,我不明白你为什么不只是使用MPI_Sendrecv或什么.你的计划试图完成什么？

UPDATE

好的,所以看起来你正在做标准的光环填充物.一些东西:

1. 每个任务都不需要它自己的数组 - 排名为0的A/AA,排名1的B/BB,等等.内存是分布式的,不是共享的; 没有排名可以看到其他阵列,所以没有必要担心覆盖它们.(如果有,则不需要发送消息).此外,想想在不同数量的进程上运行会变得多么困难 - 每次更改所使用的处理器数量时,都必须在代码中添加新数组.

2. 您可以直接读取/写入V阵列而不是使用副本,尽管最初可能最容易理解副本.

我已经写在这里使用的变量名(光环填充代码一点的版本Tmyo,Nmyo,V,indicies i和y等).每个任务只有更宽V阵列的一部分,并且仅与其邻居交换其边缘数据.它使用字符,因此您可以看到正在发生的事情.它用其排名#填充其V阵列的一部分,然后与其邻居交换其边缘数据.

我强烈建议您坐下来阅读MPI书并完成其示例.我喜欢使用MPI,但还有很多其他的.还有很多很好的MPI教程.我认为这是毫不夸张的说,MPI书籍和教程95%(例如,我们在这里 -见部分5和6)将经历正是这一程序作为其第一大合作的例子之一.他们会称之为光环填充或保护单元填充或边界交换或其他东西,但这一切都归结为通过边缘数据.

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>

char **alloc_2d_char(const int rows, const int cols) {
    char *data = (char *)malloc(rows*cols*sizeof(char));
    char **array= (char **)malloc(rows*sizeof(char*));
    for (int i=0; i<rows; i++)
        array[i] = &(data[cols*i]);

    return array;
}

void edgeDataFill(char **locV, const int locNmyo, const int locTmyo,
                  const int ncols, const int myrow, const int mycol,
                  const int size, const int rank) {

    MPI_Datatype leftright, updown;
    int left, right, up, down;
    int lefttag = 1, righttag = 2;
    int uptag = 3, downtag = 4;
    MPI_Status status;

    /* figure out our neighbours */
    left = rank-1;
    if (mycol == 0) left = MPI_PROC_NULL;

    right = rank+1;
    if (mycol == ncols-1) right = MPI_PROC_NULL;

    up = rank - ncols;
    if (myrow == 0) up = MPI_PROC_NULL;

    down = rank + ncols;
    if (down >= size) down = MPI_PROC_NULL;

    /* create data type for sending/receiving data left/right */
    MPI_Type_vector(locNmyo, 1, locTmyo+2, MPI_CHAR, &leftright);
    MPI_Type_commit(&leftright);

    /* create data type for sending/receiving data up/down */
    MPI_Type_contiguous(locTmyo, MPI_CHAR, &updown);
    MPI_Type_commit(&updown);

    /* Send edge data to our right neighbour, receive from left.
       We are sending the edge (locV[1][locTmyo]..locV[locNmyo][locTmyo]),
       and receiving into edge (locV[0][1]..locV[locNmyo][locTmyo]) */

    MPI_Sendrecv(&(locV[1][locTmyo]), 1, leftright, right, righttag,
                 &(locV[1][0]),       1, leftright, left, righttag,
                 MPI_COMM_WORLD, &status);


    /* Send edge data to our left neighbour, receive from right.
       We are sending the edge (locV[1][1]..locV[locNmyo][1]),
       and receiving into edge (locV[1][locTmyo+1]..locV[locNmyo][locTmyo+1]) */

    MPI_Sendrecv(&(locV[1][1]),         1, leftright, left,  lefttag,
                 &(locV[1][locTmyo+1]), 1, leftright, right, lefttag,
                 MPI_COMM_WORLD, &status);

    /* Send edge data to our up neighbour, receive from down.
       We are sending the edge (locV[1][1]..locV[1][locTmyo]),
       and receiving into edge (locV[locNmyo+1][1]..locV[locNmyo+1][locTmyo]) */

    MPI_Sendrecv(&(locV[1][1]),         1, updown, up,   uptag,
                 &(locV[locNmyo+1][1]), 1, updown, down, uptag,
                 MPI_COMM_WORLD, &status);

    /* Send edge data to our down neighbour, receive from up.
       We are sending the edge (locV[locNmyo][1]..locV[locNmyo][locTmyo]),
       and receiving into edge (locV[0][1]..locV[0][locTmyo]) */

    MPI_Sendrecv(&(locV[locNmyo][1]),1, updown, down, downtag,
                 &(locV[0][1]),      1, updown, up,   downtag,
                 MPI_COMM_WORLD, &status);

    /* Release the resources associated with the Type_create() calls. */

    MPI_Type_free(&updown);
    MPI_Type_free(&leftright);

}

void printArrays(char **locV, const int locNmyo, const int locTmyo,
                 const int size, const int rank) {

    /* all these barriers are a terrible idea, but it's just
       for controlling output to the screen as a demo.  You'd 
       really do something smarter here... */

    for (int task=0; task<size; task++) {
        if (rank == task) {
            printf("\nTask %d's local array:\n", rank);
            for (int i=0; i<locNmyo+2; i++) {
                putc('[', stdout);
                for (int y=0; y<locTmyo+2; y++) {
                    putc(locV[i][y], stdout);
                }
                printf("]\n");
            }
        }
        fflush(stdout);
        MPI_Barrier(MPI_COMM_WORLD);
    }
}

int main(int argc, char **argv) {
    int ierr, size, rank;
    char **locV;
    const int Nmyo=12;  /* horizontal */
    const int Tmyo=12;  /* vertical */
    const int ncols=2;  /* n procs in horizontal direction */ 
    int nrows;   
    int myrow, mycol;
    int locNmyo, locTmyo;

    ierr = MPI_Init(&argc, &argv);
    ierr|= MPI_Comm_size(MPI_COMM_WORLD, &size);
    ierr|= MPI_Comm_rank(MPI_COMM_WORLD, &rank);

    nrows = size/ncols;
    if (nrows*ncols !=  size) {
        fprintf(stderr,"Size %d does not divide number of columns %d!\n",
                size, ncols);
        MPI_Abort(MPI_COMM_WORLD,-1);
    }

    /* where are we? */
    mycol = rank % ncols;
    myrow = rank / ncols;

    /* figure out how many Tmyo we have */
    locTmyo  = (Tmyo / ncols);
    /* in case it doesn't divide evenly... */
    if (mycol == ncols-1) locTmyo = Tmyo - (ncols-1)*locTmyo;

    /* figure out how many Tmyo we have */
    locNmyo  = (Nmyo / nrows);
    /* in case it doesn't divide evenly... */
    if (myrow == nrows-1) locNmyo = Nmyo - (ncols-1)*locNmyo;

    /* allocate our local array, with space for edge data */
    locV = alloc_2d_char(locNmyo+2, locTmyo+2);

    /* fill in our local data - first spaces everywhere */
    for (int i=0; i<locNmyo+2; i++) 
        for (int y=0; y<locTmyo+2; y++) 
                locV[i][y] = ' ';

    /* then the inner regions have our rank # */
    for (int i=1; i<locNmyo+1; i++)
        for (int y=1; y<locTmyo+1; y++)
                locV[i][y] = '0' + rank;

    /* The "before" picture: */
    if (rank==0) printf("###BEFORE###\n");
    printArrays(locV, locNmyo, locTmyo, size, rank);

    /* Now do edge filling.  Ignore corners for now; 
       the right way to do that depends on your algorithm */

    edgeDataFill(locV, locNmyo, locTmyo, ncols, myrow, mycol, size, rank);

    /* The "after" picture: */
    if (rank==0) printf("###AFTER###\n");
    printArrays(locV, locNmyo, locTmyo, size, rank);

    MPI_Finalize();
}

上面的程序可以进一步简化,MPI_Cart_create用于创建你的多维域并自动为你计算你的邻居,但我想向你展示逻辑,以便你看到发生了什么.

此外,如果您能从长期以来完成此操作的人那里获得一些建议:

任何时候你都有一行一行的重复代码:像60(!!)行:

Vmax =V[i][y]-Vold; updateMaxStateChange(Vmax / dt);

mmax=m[i][y]-mold; updateMaxStateChange(mmax / dt);
hmax=h[i][y]-hold; updateMaxStateChange(hmax / dt);
jmax=j[i][y]-jold; updateMaxStateChange(jmax / dt);

mLmax=mL[i][y]-mLold; updateMaxStateChange(mLmax / dt);
hLmax=hL[i][y]-hLold; updateMaxStateChange(hLmax / dt);
hLBmax=hLB[i][y]-hLBold; updateMaxStateChange(hLBmax / dt);
hLSmax=hLS[i][y]-hLSold; updateMaxStateChange(hLSmax / dt);

amax=a[i][y]-aold; updateMaxStateChange(amax / dt);
i1fmax=i1f[i][y]-i1fold; updateMaxStateChange(i1fmax / dt);
i1smax=i1s[i][y]-i1sold; updateMaxStateChange(i1smax / dt);

Xrmax=Xr[i][y]-Xrold; updateMaxStateChange(Xrmax / dt);

i2max=i2[i][y]-i2old; updateMaxStateChange(i2max / dt);

这表明您没有使用正确的数据结构.在这里,你几乎肯定想要一个3d数组的状态变量,(可能)第三个索引是物种或本地状态变量或任何你想要调用i2,i1f,i1s等等.然后所有这些行都可以被替换使用循环,添加新的本地状态变量变得更加简单.

同样,基本上所有的状态都被定义为全局变量,这将使你的生活在更新和维护代码时变得更加困难.同样,这可能部分与以数十亿个独立状态变量的东西相关,而不是将所有相关数据组合在一起的结构或更高维数组.