Question

我正在编写一个程序来计算每个数字的总和，最多1000个。例如，1 + 2 + 3 + 4 + 5 .... + 100。首先，我将求和作业分配给10个处理器：处理器0得到1-100，处理器1得到101-200，依此类推。总和存储在一个数组中。

在所有求和并行完成后，处理器将其值发送到处理器0（处理器0使用非阻塞发送/接收来接收值），处理器0将所有值汇总并显示结果。

以下是代码：

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

using namespace std;

int summation(int, int);

int main(int argc, char ** argv)
{
    int * array;
    int total_proc;
    int curr_proc;
    int limit = 0;
    int partial_sum = 0;
    int upperlimit = 0, lowerlimit = 0;

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &total_proc);
    MPI_Comm_rank(MPI_COMM_WORLD, &curr_proc);
    MPI_Request send_request, recv_request;

    /* checking if 1000 is divisible by number of procs, else quit */
    if(1000 % total_proc != 0)
    {
        MPI_Finalize();
        if(curr_proc == 0)
            cout << "**** 1000 is not divisible by " << total_proc << " ...quitting..."<< endl;
        return 0;
    }

    /* number of partial summations */
    limit = 1000/total_proc;

    array = new int [total_proc];

    /* assigning jobs to processors */
    for(int i = 0; i < total_proc; i++)
    {
        if(curr_proc == i)
        {
            upperlimit = upperlimit + limit;
            lowerlimit = (upperlimit - limit) + 1;
            partial_sum = summation(upperlimit, lowerlimit);
            array[i] = partial_sum;
        }
        else
        {
            upperlimit = upperlimit + limit;
            lowerlimit = (upperlimit - limit) + 1;
        }
    }

    cout << "** Partial Sum From Process " << curr_proc << " is " << array[curr_proc] << endl;

    /* send and receive - non blocking */
    for(int i = 1; i < total_proc; i++)
    {
        if(curr_proc == i) /* (i = current processor) */
        {
            MPI_Isend(&array[i], 1, MPI_INT, 0, i, MPI_COMM_WORLD, &send_request);
            cout << "-> Process " << i << " sent " << array[i] << " to Process 0" << endl;

            MPI_Irecv(&array[i], 1, MPI_INT, i, i, MPI_COMM_WORLD, &recv_request);
            //cout << "<- Process 0 received " << array[i] << " from Process " << i << endl;
        }
    }

    MPI_Finalize();

    if(curr_proc == 0)
    {
        for(int i = 1; i < total_proc; i++)
            array[0] = array[0] + array[i];
        cout << "Sum is " << array[0] << endl;
    }

    return 0;
}

int summation(int u, int l)
{
    int result = 0; 
    for(int i = l; i <= u; i++)
        result = result + i;
    return result;
}

输出：

** Partial Sum From Process 0 is 5050
** Partial Sum From Process 3 is 35050
-> Process 3 sent 35050 to Process 0
<- Process 0 received 35050 from Process 3
** Partial Sum From Process 4 is 45050
-> Process 4 sent 45050 to Process 0
<- Process 0 received 45050 from Process 4
** Partial Sum From Process 5 is 55050
-> Process 5 sent 55050 to Process 0
<- Process 0 received 55050 from Process 5
** Partial Sum From Process 6 is 65050
** Partial Sum From Process 8 is 85050
-> Process 8 sent 85050 to Process 0
<- Process 0 received 85050 from Process 8
-> Process 6 sent 65050 to Process 0
** Partial Sum From Process 1 is 15050
** Partial Sum From Process 2 is 25050
-> Process 2 sent 25050 to Process 0
<- Process 0 received 25050 from Process 2
<- Process 0 received 65050 from Process 6
** Partial Sum From Process 7 is 75050
-> Process 1 sent 15050 to Process 0
<- Process 0 received 15050 from Process 1
-> Process 7 sent 75050 to Process 0
<- Process 0 received 75050 from Process 7
** Partial Sum From Process 9 is 95050
-> Process 9 sent 95050 to Process 0
<- Process 0 received 95050 from Process 9
Sum is -1544080023

打印阵列的内容：

5050
536870912
-1579286148
-268433415
501219332
32666
501222192
32666
1
0

我想知道造成这种情况的原因。

如果在调用MPI_Finalize之前打印数组，它可以正常工作。

Answer 1

你的计划最重要的缺陷就是你如何划分工作。在MPI中，每个进程都在执行主函数。因此，如果您希望它们协作构建结果，则必须确保所有进程都执行summation函数。

您不需要for循环。每个进程都分别执行main。它们只有不同的curr_proc值，您可以根据它计算他们必须执行的作业的哪个部分：

/* assigning jobs to processors */
int chunk_size = 1000 / total_proc;
lowerlimit = curr_proc * chunk_size;
upperlimit = (curr_proc+1) * chunk_size;
partial_sum = summation(upperlimit, lowerlimit);

然后，主进程如何接收所有其他进程的部分和是不正确的。

MPI排名值（curr_proc）从0开始到MPI_Comm_size输出值（total_proc-1）。
只有进程＃1正在发送/接收数据。
您使用的是发送和接收的直接版本：MPI_Isend和MPI_recv，但您不会等到这些请求完成后再等。您应该使用MPI_Waitall来实现此目的。

正确的版本如下：

if( curr_proc == 0 ) {
   // master process receives all data
   for( int i = 1; i < total_proc; i++ )
      MPI_Recv( &array[i], MPI_INT, 1, i, 0, MPI_COMM_WORLD );
} else {
   // other processes send data to the master
   MPI_Send( &partial_sum, MPI_INT, 1, 0, 0, MPI_COMM_WORLD );
}

这种一对一的通信模式称为 gather 。在MPI中，有一个功能已经执行此功能：MPI_Gather。

最后，您打算执行的操作称为 reduction ：获取给定数量的数值并通过连续执行单个操作（在您的情况下为总和）生成单个输出值。在MPI中，有一个函数也可以这样做：MPI_Reduce。

我强烈建议您在尝试制作自己之前some basic guided exercises。 MPI一开始很难理解。建立良好的基础对于您以后能够增加复杂性至关重要。 hands on tutorial也是开始使用MPI的好方法。

编辑：忘记提及您不需要按资源数量（total_proc）强制执行问题大小的均衡（在本例中为1000）。根据具体情况，您可以将余数分配给单个流程：

chunk_size = 1000 / total_proc;
if( curr_proc == 0 )
    chunk_size += 1000 % total_proc;

或者尽可能地平衡它：

int remainder = curr_proc < ( 1000 % proc )? 1 : 0;
lowerlimit = curr_proc * chunk_size /* as usual */
           + curr_proc;             /* cumulative remainder */
upperlimit = (curr_proc + 1) * chunk_size /* as usual */
           + remainder;                   /* curr_proc remainder */

第二种情况，负载不平衡将高达1，而在第一种情况下，负载不平衡在最坏的情况下可以达到total_proc-1。

Answer 2

您只是初始化array[i]，即与curr_proc ID对应的元素。该数组中的其他元素将是未初始化的，从而产生随机值。在发送/接收打印循环中，您只能访问已初始化的元素。

我不熟悉MPI，所以我猜，但你可能想在调用array之前分配MPI_Init。或者在流程0上调用MPI_Receive，而不是每个人。

MPI（总和）

2 个答案: