使用MPI_Send和MPI_Recv实现MPI_Reduce会导致错误的结果

我正在开发一个程序，它使用

MPI\u Send（）

和

MPI\u Recv（）

替换

MPI\u Reduce（）

我得到了所有要运行的东西，除了代码的最后一部分，它给出了PI近似值、错误和运行时间。我收到后也没有得到正确的和值

我相信在

MPI\u Recv（）。运行此程序时，我只使用2个处理器。当使用
MPI\u Reduce
时，如果PI未初始化为某个值，程序运行正常

#include "mpi.h"
#include <stdio.h>
#include <math.h>
 
int main( int argc, char *argv[])
{
    int n, i;
    double PI25DT = 3.141592653589793238462643;
    double pi, h, sum, x;
 
    int size, rank;
    double startTime, endTime;
 
    /* Initialize MPI and get number of processes and my number or rank*/
    MPI_Init(&argc,&argv);
    MPI_Comm_size(MPI_COMM_WORLD,&size);
    MPI_Comm_rank(MPI_COMM_WORLD,&rank);
 
    /* Processor zero sets the number of intervals and starts its clock*/
    if (rank==0)
    {
       n=600000000;
       startTime=MPI_Wtime();
       for (int i = 0; i < size; i++) {
           if (i != rank) {
               MPI_Send(&n, 1, MPI_INT, i, 0, MPI_COMM_WORLD);
           }
       }
    } 
    /* Broadcast number of intervals to all processes */
    else 
    {
        MPI_Recv(&n, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
    }
 
    /* Calculate the width of intervals */
    h   = 1.0 / (double) n;
 
    /* Initialize sum */
    sum = 0.0;
    /* Step over each inteval I own */
    for (i = rank+1; i <= n; i += size)
    {
        /* Calculate midpoint of interval */
        x = h * ((double)i - 0.5);
        /* Add rectangle's area = height*width = f(x)*h */
        sum += (4.0/(1.0+x*x))*h;
    }
    /* Get sum total on processor zero */
    //MPI_Reduce(&sum,&pi,1,MPI_DOUBLE,MPI_SUM,0,MPI_COMM_WORLD);
    MPI_Send(&sum, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD);
    MPI_Send(&pi, 1, MPI_SUM, 0, 0, MPI_COMM_WORLD);
    
    if (rank == 0) 
    {
        double total_sum = 0;
        for (int i = 0; i < size; i++) 
        {
            MPI_Recv(&sum, 1, MPI_DOUBLE, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
            MPI_Recv(&pi, 1, MPI_SUM, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
            total_sum += sum;
            printf("Total Sum is %lf\n", total_sum);
        }
    }
    
    /* Print approximate value of pi and runtime*/
    if (rank==0)
    {
       printf("pi is approximately %.16f, Error is %e\n",
                       pi, fabs(pi - PI25DT));
       endTime=MPI_Wtime();
       printf("runtime is=%.16f",endTime-startTime);
    }
    MPI_Finalize();
    return 0;
}

#包括“mpi.h”
#包括
#包括
int main（int argc，char*argv[]）
{
int n，i；
双PI25DT=3.141592653589793238462643；
双π，h，和，x；
整数大小、等级；
双开始时间，结束时间；
/*初始化MPI并获取进程数和我的编号或排名*/
MPI_Init（&argc，&argv）；
MPI_通信大小（MPI_通信世界和大小）；
MPI通信等级（MPI通信世界和等级）；
/*处理器zero设置间隔数并启动其时钟*/
如果（秩==0）
{
n=600000000；
startTime=MPI_Wtime（）；
对于（int i=0；i
及

错误，应作为第三个参数
MPI\u数据类型
而不是
MPI\u OP
（即
MPI\u总和
）

但看看你的代码，你真正想做的是用以下方法替换这些调用：

double pi = sum;
if (myid != 0) {
        MPI_Send(&sum, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD);
}
else {
    for (int i = 1; i < numprocs; i++) {
        MPI_Recv(&value, 1, MPI_DOUBLE, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
        pi += value;
        }
}

输出（2个过程）：

 MPI_Recv(&pi, 1, MPI_SUM, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
                  ^^^^^^^

double pi = sum;
if (myid != 0) {
        MPI_Send(&sum, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD);
}
else {
    for (int i = 1; i < numprocs; i++) {
        MPI_Recv(&value, 1, MPI_DOUBLE, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
        pi += value;
        }
}

#include "mpi.h"
#include <stdio.h>
#include <math.h>
 
int main( int argc, char *argv[])
{
    int n, i;
    double PI25DT = 3.141592653589793238462643;
    double h, sum, x;
 
    int numprocs, myid;
    double startTime, endTime;
 
    /* Initialize MPI and get number of processes and my number or rank*/
    MPI_Init(&argc,&argv);
    MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
    MPI_Comm_rank(MPI_COMM_WORLD,&myid);
 
    /* Processor zero sets the number of intervals and starts its clock*/
    if (myid==0) {
       n=600000000;
       startTime=MPI_Wtime();
       for (int i = 0; i < numprocs; i++) {
           if (i != myid) {
               MPI_Send(&n, 1, MPI_INT, i, 0, MPI_COMM_WORLD);
           }
       }
    } 
    else {
        MPI_Recv(&n, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
    }
 
    /* Calculate the width of intervals */
    h   = 1.0 / (double) n;
 
    /* Initialize sum */
    sum = 0.0;
    /* Step over each inteval I own */
    for (i = myid+1; i <= n; i += numprocs) {
        /* Calculate midpoint of interval */
        x = h * ((double)i - 0.5);
        /* Add rectangle's area = height*width = f(x)*h */
        sum += (4.0/(1.0+x*x))*h;
    }
    /* Get sum total on processor zero */
    //MPI_Reduce(&sum,&pi,1,MPI_DOUBLE,MPI_SUM,0,MPI_COMM_WORLD);
    double value = 0;
    double pi = sum;
    if (myid != 0) {
            MPI_Send(&sum, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD);
    }
    else {
        for (int i = 1; i < numprocs; i++) {
            MPI_Recv(&value, 1, MPI_DOUBLE, i, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
            pi += value;
            }
    }
    
    /* Print approximate value of pi and runtime*/
    if (myid==0) {
       printf("pi is approximately %.16f, Error is %e\n",
                       pi, fabs(pi - PI25DT));
       endTime=MPI_Wtime();
       printf("runtime is=%.16f",endTime-startTime);
    }
    MPI_Finalize();
    return 0;
}

pi is approximately 3.1415926535898993, Error is 1.061373e-13
runtime is=1.3594319820404053