MPI c++；环形拓扑发送和接收不同的值，而只传递相同的值？ 我只是学习C++中的环拓扑MPI。我编写了一个C++脚本，计算10维Monte积分，并计算其平均值和局部最大值。我的目标是通过“环”传递每个proessors的本地最大值

现在，我仍然不知道如何在一个数组中存储运行时从不同处理器生成的最大值，所以我编译并执行了一次代码，并用这些值手动创建了一个数组

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。 现在我正在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++是否使用MPI库，我遵循了C++的MPI在线教程，并且使用了与C++代码中的C结构相同的结构。p> 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;
}

当然，输出的平均值与amd max不同，但我现在担心send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

---

<>我想我在C和C++中使用MPI的时候，我会很感激任何建议，而且我没有看到任何优秀的C++ MPI教程，所以我的代码或教程链接的一个很好的修改例子会很有帮助。谢谢

MPI_Recv和

MPI_Send

的第三个参数是数据类型。现在您正在发送一个

double

，但是您将数据类型设置为

MPI\u INT

。在大多数系统中，

int

是4个字节，

double

是8个字节，因此，

rec\u junk

中有一半字节未初始化

要修复它，只需在所有调用

MPI\u Recv

和

MPI\u Send

时将

MPI\u INT

更改为

MPI\u DOUBLE

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