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();
}

任何人都能解释一下上面的代码是什么吗？我在实验室，我的思念让我解释，我不明白。

Answer 1

这是使用MPICH库的并行处理来计算pi的示例。此示例包含在mpich安装中。它基本上通过在多个处理器或主机之间添加分数来划分计算pi的工作。

代码说明（MPICH）

1 个答案: