用MPI进行分布式内存编程(1)

《并行程序设计导论》第三章部分程序

程序3.1运行实例

#include<stdio.h>
#include<string.h>
#include<mpi.h>

const int MAX_STRING=100;

int main()
{
    char greet[MAX_STRING];
    int comm_sz; //进程数
    int my_rank; //进程号
    
    MPI_Init(NULL,NULL); //初始化
    MPI_Comm_size(MPI_COMM_WORLD,&comm_sz); //返回进程数 
    MPI_Comm_rank(MPI_COMM_WORLD,&my_rank); //返回进程号
    
    if(my_rank!=0)
    {
        sprintf(greet,"Greeting from process %d of %d",my_rank,comm_sz);
        MPI_Send(greet,strlen(greet)+1,MPI_CHAR,0,0,MPI_COMM_WORLD); //通信,发送
    }
    else
    {
        printf("Greeting from process %d of %d",my_rank,comm_sz);
        int q;
        for( q=1;q<comm_sz;q++)
        {
            MPI_Recv(greet,MAX_STRING,MPI_CHAR,q,0,MPI_COMM_WORLD,MPI_STATUS_IGNORE);//通信,接收
            printf("%s
",greet);
        }
    }
    MPI_Finalize(); //告知系统MPI已使用完毕
    return 0;
    
    
}

在天河平台运行结果

 自己虚拟机运行结果

3.2运行实例

#include<stdio.h>
#include<string.h>
#include<mpi.h>

double f(double x)
{
    return x*x+x*x*x+5;
}
double Trap(double left_endpt,double right_endpt,int trap_count,double base_len)
{
    
    
    double estimate,x;
    int i;
    estimate=(f(left_endpt)+f(right_endpt))/2.0; //梯形面积
    for(i =1;i<=trap_count-1;++i)
    {
        x=left_endpt+i*base_len;
        estimate+=f(x);
    }
    estimate=estimate*base_len;
    return estimate;
}

int main()
{
    int my_rank,comm_sz,n=1024,local_n;
    double a=0.0,b=3.0,h,local_a,local_b;
    double local_int,total_int;
    int source;
    
    MPI_Init(NULL,NULL); //初始化
    MPI_Comm_size(MPI_COMM_WORLD,&comm_sz); //返回进程数
    MPI_Comm_rank(MPI_COMM_WORLD,&my_rank); //返回进程号
    
    h=(b-a)/n;
    local_n=n/comm_sz;
    
    local_a=a+my_rank*local_n*h;
    local_b=local_a+local_n*h;
    local_int=Trap(local_a,local_b,local_n,h);
    
    if(my_rank!=0)
    {
        MPI_Send(&local_int,1,MPI_DOUBLE,0,0,MPI_COMM_WORLD);
    }
    else
    {
        total_int=local_int;
        for(source=1;source<comm_sz;source++)
        {
            MPI_Recv(&local_int,1,MPI_DOUBLE,source,0,MPI_COMM_WORLD,MPI_STATUS_IGNORE); //接受其他节点信息
            total_int+=local_int;
        }
    }
    
    if(my_rank==0)
    {
        printf("With n=%d trapezoids,our estimated
",n);
        printf("of the intergral from %f to %f=%.15e
",a,b,total_int);
    }
    
    MPI_Finalize();
    return 0;
    
    
    
    
    
    
    
    
}

天河运行结果

原文地址:https://www.cnblogs.com/caishunzhe/p/12935439.html