Question

我刚学习c ++中的环形拓扑MPI。我写了一个c ++脚本来计算10维蒙特卡罗积分并计算其平均值和局部最大值。我的目标是通过“响铃”传递每个教授的本地最大值。

现在，我仍然没有弄清楚如何在运行时将不同处理器生成的最大值存储在一个数组中，因此我编译并执行了一次代码，并手动创建了一个包含值的数组。

接下来我想通过环传递每个数组值，并最终计算全局最大值。现在我只是尝试传递第一个数组值，我看到处理器发送相同的值但接收不同的值。老实说，我不知道c ++是否以不同的方式使用MPI库，我遵循了一个带有C的MPI在线教程，并且使用了与C在我的c ++代码中相同的结构。

我在这里分享代码。

#include <iostream>
#include <fstream>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <mpi.h>
using namespace std;


//define multivariate function F(x1, x2, ...xk)            

double f(double x[], int n)
{
    double y;
    int j;
    y = 0.0;

    for (j = 0; j < n-1; j = j+1)
      {
         y = y + exp(-pow((1-x[j]),2)-100*(pow((x[j+1] - pow(x[j],2)),2)));

      }     

    y = y;
    return y;
}

//define function for Monte Carlo Multidimensional integration

double int_mcnd(double(*fn)(double[],int),double a[], double b[], int n, int m)

{
    double r, x[n], v;
    int i, j;
    r = 0.0;
    v = 1.0;
    // initial seed value (use system time) 
    //srand(time(NULL)); 


    // step 1: calculate the common factor V
    for (j = 0; j < n; j = j+1)
      {
         v = v*(b[j]-a[j]);
      } 

    // step 2: integration
    for (i = 1; i <= m; i=i+1)
    {
        // calculate random x[] points
        for (j = 0; j < n; j = j+1)
        {
            x[j] = a[j] +  (rand()) /( (RAND_MAX/(b[j]-a[j])));
        }         
        r = r + fn(x,n);
    }
    r = r*v/m;

    return r;
}




double f(double[], int);
double int_mcnd(double(*)(double[],int), double[], double[], int, int); 



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

    int rank, size;

    MPI_Init (&argc, &argv);      // initializes MPI
    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));

        if( mean > max) 
        {
         max = mean;
        }
        //cout << setw(10)  << m << setw(10) << max << setw(10) << mean << setw(10) << rank << setw(10) << size <<endl;
        m = m*4; 
    }

    //cout << setw(30)  << m << setw(30) << result << setw(30) << mean <<endl; 
    printf("Process %d of %d mean = %1.5e\n and local max = %1.5e\n", rank, size, mean, max );


    double max_store[4] = {4.43095e-02, 5.76586e-02, 3.15962e-02, 4.23079e-02}; 

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


  // This next if-statment implemeents the ring topology
  // the last process ID is size-1, so the ring topology is: 0->1, 1->2, ... size-1->0
  // rank 0 starts the chain of events by passing to rank 1
  if(rank==0) {
    // only the process with rank ID = 0 will be in this block of code.
    MPI_Send(&send_junk, 1, MPI_INT, 1, 0, MPI_COMM_WORLD); //  send data to process 1
    MPI_Recv(&rec_junk, 1, MPI_INT, size-1, 0, MPI_COMM_WORLD, &status); // receive data from process size-1
  }
  else if( rank == size-1) { 
    MPI_Recv(&rec_junk, 1, MPI_INT, rank-1, 0, MPI_COMM_WORLD, &status); // recieve data from process rank-1 (it "left" neighbor")
    MPI_Send(&send_junk, 1, MPI_INT, 0, 0, MPI_COMM_WORLD); // send data to its "right neighbor", rank 0
  }
  else {
    MPI_Recv(&rec_junk, 1, MPI_INT, rank-1, 0, MPI_COMM_WORLD, &status); // recieve data from process rank-1 (it "left" neighbor")
    MPI_Send(&send_junk, 1, MPI_INT, rank+1, 0, MPI_COMM_WORLD); // send data to its "right neighbor" (rank+1)
  }
  printf("Process %d send %1.5e\n and recieved %1.5e\n", rank, send_junk, rec_junk ); 


  MPI_Finalize(); // programs should always perform a "graceful" shutdown
    return 0;
}

我编译：

mpiCC -std=c++11 -o hg test_code.cpp
mpirun -np 4 ./hg

输出看起来像这样，当然amd max不同，但我现在担心发送和recvd值：

Process 2 of 4 mean = 2.81817e-02
 and local max = 5.61707e-02
Process 0 of 4 mean = 2.59220e-02
 and local max = 4.43095e-02
Process 3 of 4 mean = 2.21734e-02
 and local max = 4.30539e-02
Process 1 of 4 mean = 2.87403e-02
 and local max = 6.58530e-02
Process 1 send 4.43095e-02
 and recieved 2.22181e-315
Process 2 send 4.43095e-02
 and recieved 6.90945e-310
Process 3 send 4.43095e-02
 and recieved 6.93704e-310
Process 0 send 4.43095e-02
 and recieved 6.89842e-310

我认为我搞砸了C和c ++中的MPI用法，我很感激任何建议，我也没有看到任何关于互联网的好的c ++ MPI教程，所以我的代码或教程链接的一个很好的修改示例将会非常有帮助。感谢

Answer 1

MPI_Recv和MPI_Send的第三个参数是数据类型。现在您要发送double，但是您将数据类型设置为MPI_INT。在大多数系统中，int为4个字节，double为8个字节，因此rec_junk中的一半字节未初始化。

要解决此问题，只需在MPI_INT和MPI_DOUBLE的所有来电中将MPI_Recv更改为MPI_Send。

MPI c ++环拓扑发送和接收不同的值，同时只传递相同的值？

1 个答案: