Question

您好我希望找到所有进程的局部最大值，然后将所有局部最大值传递给所有进程以生成单个数组，然后使用MPI环拓扑来比较局部最大值，然后最终输出全局最大值。

我可以更高效地执行MPI_ALLreduce并且已经这样做了，但我想测试环形拓扑的效率并产生与allreduce相同的结果。我正在按照教程使用MPI_Allgather，它返回了一些错误。代码如下：

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

    int rank, size;

    MPI_Init (&argc, &argv);      // initializes MPI
    //MPI_Comm comm;
    double max_store[4];
    double *rbuf;
    //MPI_Comm_size( comm, &rank);
    MPI_Comm_rank (MPI_COMM_WORLD, &rank); // get current MPI-process ID. O, 1, ...
    MPI_Comm_size (MPI_COMM_WORLD, &size); // get the total number of processes


    /* define how many integrals */
    const int n = 10;       

    double b[n] = {5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0,5.0};                    
    double a[n] = {-5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0,-5.0};  

    double result, mean;
    int m;

    const unsigned int N = 5;
    double max = -1;



    cout.precision(6);
    cout.setf(ios::fixed | ios::showpoint); 


    srand(time(NULL) * rank);  // each MPI process gets a unique seed

    m = 4;                // initial number of intervals

    // convert command-line input to N = number of points
    //N = atoi( argv[1] );


    for (unsigned int  i=0; i <=N; i++)
    {
        result = int_mcnd(f, a, b, n, m);
        mean = result/(pow(10,10));

        m = m*4; 
        if( mean > max) 
        {
         max = mean;
        }


    }
    //if ( rank < 4 && rank >= 0 ) 
        //{
         //max_store[rank] = max;
        //}





    printf("Process ID %i, local_max = %f\n",rank, max);


    // All processes get the global max, stored in place of the local max
    MPI_Allreduce( MPI_IN_PLACE, &max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD );

    printf("Process ID %d, global_max = %f\n",rank, max);

    rbuf = (double *)malloc(rank*4*sizeof(double)); 
    MPI_Allgather( max_store, 4, MPI_DOUBLE, rbuf, 4, MPI_DOUBLE, MPI_COMM_WORLD);
    //print the array containing max from each processor
    //int k;    
    //for( int k = 0; k < 4; k++ )
    //{
     //printf( "%1.5e\n", max_store[k]);
    //}



   double send_junk = max_store[0];
   double rec_junk;
    //double global_max;
    MPI_Status status;

  if(rank==0) 
  {
    MPI_Send(&send_junk, 4, MPI_DOUBLE, 1, 0, MPI_COMM_WORLD); //  send data to process 1
  }
  if(rank==1) 
  {
    MPI_Recv(&rec_junk, 4, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD, &status); // receive data from process 0
  }
  //check between process 0 and process 1 maxima
  if(rec_junk>=max_store[1])
  {
   rec_junk = max_store[0];
  }
  else
  {
   rec_junk = max_store[1];
  }
  send_junk = rec_junk;

  MPI_Send(&send_junk, 4, MPI_DOUBLE, 2, 0, MPI_COMM_WORLD); //  send data to process 2

  if(rank==2)
  {
   MPI_Recv(&rec_junk, 4, MPI_DOUBLE, 1, 0, MPI_COMM_WORLD, &status); // receive data from process 1
  }
  //check between process 1 and process 2 maxima
  if(rec_junk>=max_store[2])
  {
   rec_junk = rec_junk;
  }
  else
  {
   rec_junk = max_store[2];
  }
  send_junk = rec_junk;

  MPI_Send(&send_junk, 4, MPI_DOUBLE, 3, 0, MPI_COMM_WORLD); //  send data to process 3

  if(rank==3)
  {
   MPI_Recv(&rec_junk, 4, MPI_DOUBLE, 2, 0, MPI_COMM_WORLD, &status); // receive data from process 2
  }
  //check between process 2 and process 3 maxima 
  if(rec_junk>=max_store[3])
  {
   rec_junk = rec_junk;
  }
  else
  {
   rec_junk = max_store[3];
  }


  printf("global ring max = %f", rec_junk); 


  MPI_Finalize(); // programs should always perform a "graceful" shutdown

  return 0;
}

我很想知道如何在单个阵列中发送最大值，并且所有进程都可以访问它，这样我就可以比较环形拓扑中的值。非常感谢。

Answer 1

您没有正确分配接收缓冲区。它需要足够大才能存储每个等级的4个条目。你现在有：

rbuf = (double *)malloc(rank*4*sizeof(double));

什么时候应该

rbuf = (double *)malloc(size*4*sizeof(double));

MPI_ALLgather将局部最大值发送到所有进程，不良终止进程

1 个答案: