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我正在学习 MPI-2/MPI-3 中引入的 MPI 单面通信，并遇到了这个在线课程页面：MPI_Accumulate

MPI_Accumulate 允许调用者将移动到目标进程的数据与已经存在的数据结合起来，例如在目标进程中累积总和。通过使用 MPI_Get 检索数据（随后是同步）可以实现相同的功能；在调用者处执行求和运算；然后使用 MPI_Put 将更新的数据发送回目标进程。积累简化了这种混乱......

但是只有有限数量的操作允许与MPI_Accumulate（max、min、sum、product 等）一起使用，并且不允许用户定义的操作。我想知道如何实现上面提到的混乱，使用MPI_Get、同步、操作和MPI_Put. 是否有任何 C/C++ 教程或工作代码示例？

谢谢

为了测试，我改编了这个SO question中的一段代码，其中使用单面通信来创建一个整数计数器，该计数器在 MPI 进程中保持同步。使用的目标问题行MPI_Accumulate已标记。

代码按原样编译并在大约 15 秒内返回。但是当我尝试用MPI_Accumulate问题行之后的注释块中所示的等效基本操作序列替换时，编译的程序无限期挂起。

谁能帮忙解释一下出了什么问题，MPI_Accumulate在这种情况下正确的替换方法是什么？

PS我编译了代码

g++ -std=c++11 -I..   mpistest.cpp -lmpi

并执行二进制文件

mpiexec -n 4 a.exe

代码：

//adpated from https://stackoverflow.com/questions/4948788/
#include <mpi.h>
#include <stdlib.h>
#include <stdio.h>
#include <thread>
#include <chrono>

struct mpi_counter_t {
  MPI_Win win;
  int  hostrank;  //id of the process that host values to be exposed to all processes
  int  rank;    //process id
  int  size;     //number of processes
  int  val;      
  int  *hostvals; 
};

struct mpi_counter_t *create_counter(int hostrank) {
    struct mpi_counter_t *count;

    count = (struct mpi_counter_t *)malloc(sizeof(struct mpi_counter_t));
    count->hostrank = hostrank;
    MPI_Comm_rank(MPI_COMM_WORLD, &(count->rank));
    MPI_Comm_size(MPI_COMM_WORLD, &(count->size));

    if (count->rank == hostrank) {
        MPI_Alloc_mem(count->size * sizeof(int), MPI_INFO_NULL, &(count->hostvals));
        for (int i=0; i<count->size; i++) count->hostvals[i] = 0;
        MPI_Win_create(count->hostvals, count->size * sizeof(int), sizeof(int),
                       MPI_INFO_NULL, MPI_COMM_WORLD, &(count->win));
    } 
    else {
        count->hostvals = NULL;
        MPI_Win_create(count->hostvals, 0, 1,
                       MPI_INFO_NULL, MPI_COMM_WORLD, &(count->win));
    }
    count -> val = 0;

    return count;
}

int increment_counter(struct mpi_counter_t *count, int increment) {
    int *vals = (int *)malloc( count->size * sizeof(int) );
    int val;

    MPI_Win_lock(MPI_LOCK_EXCLUSIVE, count->hostrank, 0, count->win);

    for (int i=0; i<count->size; i++) {

        if (i == count->rank) {
        MPI_Accumulate(&increment, 1, MPI_INT, 0, i, 1, MPI_INT, MPI_SUM,count->win); //Problem line: increment hostvals[i] on host
        /* //Question: How to correctly replace the above MPI_Accumulate call with the following sequence? Currently, the following causes the program to hang.
            MPI_Get(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
            MPI_Win_fence(0,count->win);
            vals[i] += increment;
            MPI_Put(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
            MPI_Win_fence(0,count->win);
        //*/
        } else {
            MPI_Get(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
        }
    }

    MPI_Win_unlock(0, count->win);

    //do op part of MPI_Accumulate's work on count->rank
    count->val += increment; 
    vals[count->rank] = count->val; 

    //return the sum of vals
    val = 0;
    for (int i=0; i<count->size; i++)
        val += vals[i];

    free(vals);
    return val;
}

void delete_counter(struct mpi_counter_t **count) {
    if ((*count)->rank == (*count)->hostrank) {
        MPI_Free_mem((*count)->hostvals);
    }
    MPI_Win_free(&((*count)->win));
    free((*count));
    *count = NULL;

    return;
}

void print_counter(struct mpi_counter_t *count) {
    if (count->rank == count->hostrank) {
        for (int i=0; i<count->size; i++) {
            printf("%2d ", count->hostvals[i]);
        }
        puts("");
    }
}


int main(int argc, char **argv) {

    MPI_Init(&argc, &argv);

    const int WORKITEMS=50;

    struct mpi_counter_t *c;
    int rank;
    int result = 0;

    c = create_counter(0);

    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    srand(rank);

    while (result < WORKITEMS) {
        result = increment_counter(c, 1);
        if (result <= WORKITEMS) {
             printf("%d working on item %d...\n", rank, result);
         std::this_thread::sleep_for (std::chrono::seconds(rand()%2));
         } else {
             printf("%d done\n", rank);
         }
    }

    MPI_Barrier(MPI_COMM_WORLD);
    print_counter(c);
    delete_counter(&c);


    MPI_Finalize();
    return 0;
}

还有一个问题，我应该MPI_Win_fence在这里使用而不是锁吗？

- 编辑 -

increment_counter我按如下方式使用锁定/解锁，程序运行但行为奇怪。在最终的打印输出中，主节点完成所有工作。还是一头雾水。

int increment_counter(struct mpi_counter_t *count, int increment) {
    int *vals = (int *)malloc( count->size * sizeof(int) );
    int val;

    MPI_Win_lock(MPI_LOCK_EXCLUSIVE, count->hostrank, 0, count->win);

    for (int i=0; i<count->size; i++) {

        if (i == count->rank) {
            //MPI_Accumulate(&increment, 1, MPI_INT, 0, i, 1, MPI_INT, MPI_SUM,count->win); //Problem line: increment hostvals[i] on host
            ///* //Question: How to correctly replace the above MPI_Accumulate call with the following sequence? reports that 0 does all the work
            MPI_Get(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
            MPI_Win_unlock(0, count->win);
            vals[i] += increment;
            MPI_Put(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
            MPI_Win_lock(MPI_LOCK_EXCLUSIVE, count->hostrank, 0, count->win);
        //*/
        } else {
            MPI_Get(&vals[i], 1, MPI_INT, 0, i, 1, MPI_INT, count->win);
        }
    }

    MPI_Win_unlock(0, count->win);

    //do op part of MPI_Accumulate's work on count->rank
    count->val += increment; 
    vals[count->rank] = count->val; 

    //return the sum of vals
    val = 0;
    for (int i=0; i<count->size; i++)
        val += vals[i];

    free(vals);
    return val;
}
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1 回答 1

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使用 Gets 和 Puts 实现 Accumulate 确实会非常麻烦，尤其是当您必须处理派生数据类型等时。但是假设您正在对单个整数进行累积，并且您只想将本地值汇总到远程缓冲区中，您可以执行以下操作（仅限伪代码）：

MPI_Win_lock(EXCLUSIVE);  /* exclusive needed for accumulate atomicity constraints */
MPI_Get(&remote_data);
MPI_Win_flush(win);  /* make sure GET has completed */
new = local_data + remote_data;
MPI_Put(&new);
MPI_Win_unlock();

您的代码不正确，因为您在 GET 之后放弃了排他锁，这会在两个进程同时尝试对数据求和时导致原子性问题。

于 2014-08-26T03:24:36.287 回答