Example-1-MPI_Scatterv and MPI_Gather

目的：利用例子说明collective communication函数功能

Example 1: This program shows how to use MPI_Scatterv and MPI_Gather. 假设有np个数据，要分配到nprocs个进程中，最后一个进程得到平均分配之后多余的数据。每个进程对接收到的数据进行一些改变，然后求和，最后master进程收集求和结果并打印输出。

#include "mpi.h"
#include <cstdio>
#include <math.h>
#include <iostream>

#define np 105  //number of elements need to be scattered at root
const int MASTER = 0; // Rank of of the master process

int main(int argc, char* argv[])
{
double PosX[np]; //elements need to be scattered
double *RecPosX; //receive buffer
int nProcs , Rank ;//number of processes, and id of rank
int i,j;
int reminder; //reminder if it is not divisible
double StartTime,EndTime; //timing

MPI_Init ( & argc , & argv );
MPI_Comm_size ( MPI_COMM_WORLD , & nProcs );
MPI_Comm_rank ( MPI_COMM_WORLD , & Rank );

StartTime = MPI_Wtime(); //start timing

//initialize array
int* sendcounts = NULL ;
int* displs = NULL ;

if ( Rank == MASTER )
{
for(i=0;i<np;i++)
{
PosX[i] = i*1.0; //initialize elements in master process
}
}

sendcounts = new int [ nProcs ]; //allocate memory for array storing number of elements
reminder = np%nProcs; //remineder

//calculate number of elements need to be scattered in each process
for(i=0;i<nProcs;i++)
{
sendcounts [i] = int(1.0*np/nProcs);
}
//number of elements in the last process
sendcounts [nProcs-1] = sendcounts [nProcs-1] + reminder;
//calculate corresponding displacement

displs = new int [ nProcs ];
for(i=0;i<nProcs;i++)
{
displs[i] = 0;
for(j=0;j<i;j++)
{
displs[i] = displs[i] + sendcounts[j];
}
}
//allocate the receive buffer
RecPosX = (double*)malloc(sendcounts [Rank]*sizeof(double));
//now everything is ready and we can start MPI_Scatterv operation.
MPI_Scatterv(PosX, sendcounts , displs, MPI_DOUBLE,
RecPosX, sendcounts[Rank], MPI_DOUBLE, 0, MPI_COMM_WORLD);

//output results after MPI_Scatterv operation
//std::cout<<"My Rank = "<<Rank<<" And data I received are:"<<std::endl;
double sum=0;
double *rbuf;

rbuf=(double*)malloc(nProcs*sizeof(double));

for(i=0;i<sendcounts[Rank];i++)
{
RecPosX[i] = RecPosX[i] + Rank/10.0;
sum += RecPosX[i];
//std::cout<<RecPosX[i]<<" ";
//if(i%10 == 0 && i != 0)std::cout<<std::endl;
}
std::cout<<std::endl;

MPI_Gather(&sum,1,MPI_DOUBLE,rbuf,1,MPI_DOUBLE,MASTER,MPI_COMM_WORLD);

EndTime = MPI_Wtime();
if(Rank == MASTER)
{
for(i=0;i<nProcs;i++)
{
std::cout<<"Rank = "<<i<<" Sum= "<<rbuf[i]<<std::endl;
}
std::cout<<"Total Time Spending For: "<<nProcs<<" Is "<<EndTime-StartTime<<" :Second"<<std::endl;
}

MPI_Finalize();
return 0;

}

Assuming np=105, nprocs=4. then the output will be like these:

Rank=0 Sum=325

Rank=1 Sum=1003.6

Rank=2 Sum=1682.6

Rank=3 Sum=2465.1