代码解释（MPICH）

Question

#include "mpi.h" 
#include <stdio.h> 
#include <math.h> 


double f(double a) 
{ 
    return (4.0/(1.0 + a*a)); 
} 

void main(int argc, char *argv[]) 
{ 
    int done = 0, n, myid, numprocs,i; 
    double PI25DT = 3.141592653589793238462643; 
    double mypi, pi, h, sum, x; 
    double startwtime, endwtime; 
    int namelen; 
    char processor_name[MPI_MAX_PROCESSOR_NAME];  
    MPI_Init(&argc,&argv); 
    MPI_Comm_size(MPI_COMM_WORLD,&numprocs); 
    MPI_Comm_rank(MPI_COMM_WORLD,&myid); 
    MPI_Get_processor_name(processor_name,&namelen);  
    fprintf(stderr,"Process %d on %s\n", 
     myid, processor_name); 
    fflush(stderr); 
    n = 0; 
    while (!done) 
    { 
     if (myid == 0) 
     { 
     printf("Enter the number of intervals: (0 quits) ");fflush(stdout); 
     scanf("%d",&n); 

     startwtime = MPI_Wtime(); 
     } 
     MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD); 
     if (n == 0) 
      done = 1; 
     else 
     { 
      h = 1.0/(double) n; 
      sum = 0.0; 
      for (i = myid + 1; i <= n; i += numprocs) 
      { 
       x = h * ((double)i - 0.5); 
       sum += f(x); 
      } 
      mypi = h * sum; 

      MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD); 

      if (myid == 0) 
     { 
       printf("pi is approximately %.16f, Error is %.16f\n", 
      pi, fabs(pi - PI25DT)); 
     endwtime = MPI_Wtime(); 
     printf("wall clock time = %f\n", endwtime-startwtime);   
     } 
     } 
    } 
    MPI_Finalize(); 
} 

任何人都可以解释我上面的代码是什么吗？我在实验室，我的小姐让我解释，我不明白。

Answer 1

这是使用MPICH库的平行处理来计算pi的例子。该示例包含在mpich安装中。它本质上是通过在多个处理器或主机之间添加分数来分割计算pi的工作。