如何在 C++ 中使用 MPI 同步和排序打印（任务）
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【中文标题】如何在 C++ 中使用 MPI 同步和排序打印（任务）【英文标题】：How to synchronize and sort prints (tasks) with MPI in C++ 【发布时间】：2021-10-17 07:39:57 【问题描述】：编辑前：
我正在使用 MPI 用 C++ 编写一个简单的代码来进行并行处理。我做一些简单的任务，比如从一个进程向另一个进程发送和接收消息，与MPI_Sendrecv 交换消息，打个招呼，并打印每个进程的执行时间。它可以工作，但输出没有按我想要的顺序排序（进程 0：任务......进程 1：任务......）。
我知道这是由于进程之间没有同步，根据我的搜索，我知道 MPI_Send,MPI_Recv ... 是隐式阻塞函数，但似乎我不明白如何使用这个功能。我还尝试了经常推荐的 MPI_Barrier() 功能，但没有成功。我运行 8 个进程。
也许有人可以帮助我？提前致谢。
这是我的代码：
#include <mpi.h>
#include <iostream>
using namespace std;
int main(int argc, char* argv[])
int rank,nbproc, length;
char name[80];
float time;
double SendData, ReceiveData;
int tag = 1;
int NumTo6 = 500;
int NumTo7 = 300;
int ReceiveFrom6, ReceiveFrom7;
char message[] = "pokay";
char receive[] = "none";
int longueur = strlen(message);
SendData = 1254.3356;
ReceiveData = 0;
MPI_Init(&argc, &argv);
time = MPI_Wtime();
cout << " " << endl;
MPI_Comm_size(MPI_COMM_WORLD, &nbproc);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
// MPI_Get_processor_name(name, &length);
cout << "Hello from process\t" << rank << endl;
if (rank == 1)
cout << "2*5 = " << 2*5 << endl;
if (rank == 2)
MPI_Send(&SendData,1,MPI_DOUBLE,3,tag,MPI_COMM_WORLD);
if (rank == 3)
cout << "Data before the reception:\t" << ReceiveData << endl;
MPI_Recv(&ReceiveData,1,MPI_DOUBLE,2,tag,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << "Data received is :\t" << ReceiveData << endl;
tag+=1;
if (rank == 4)
MPI_Send(&message[1],4,MPI_CHAR,5,tag,MPI_COMM_WORLD);
if (rank == 5)
cout << "Message before the reception:\t" << receive << endl;
MPI_Recv(&receive,longueur+1,MPI_CHAR,4,tag,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << "Message after reception:\t" << receive << endl;
tag+=1;
// Exchange between 2 processes:
if(rank == 6)
MPI_Sendrecv(&NumTo7,1,MPI_INT,7,tag,&ReceiveFrom7,1,MPI_INT,7,tag+1,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << "Num receive from 7:\t" << ReceiveFrom7 << endl;
if(rank == 7)
MPI_Sendrecv(&NumTo6,1,MPI_INT,6,tag+1,&ReceiveFrom6,1,MPI_INT,6,tag,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << "Num receive from 6:\t" << ReceiveFrom6 << endl;
tag+=2;
time = MPI_Wtime() - time;
cout << "Time spend on process " << rank << " is: " << time << " sec" << endl;
MPI_Finalize();
return 0;
这是我的输出：
Hello from process 6
Hello from process 7
Num receive from 6: 300
Time spend on process 7 is: 6.0746e-05 sec
Hello from process 2
Time spend on process 2 is: 5.0439e-05 sec
Hello from process 3
Data before the reception: 0
Data received is: 1254.34
Time spend on process 3 is: 0.000439355 sec
Hello from process 4
Time spend on process 4 is: 6.2342e-05 sec
Hello from process 5
Message before the reception: none
Message after reception: okay
Time spend on process 5 is: 0.000168845 sec
Hello from process 1
2*5 = 10
Time spend on process 1 is: 0.000132448 sec
Hello from process 0
Time spend on the process 0 is: 3.9762e-05 sec
Num receive from 7: 500
Time spend on process 6 is: 0.00206599 sec
编辑 @VictorEijkhout 的评论：
除了char 的 MPI_Gather() 之外，我几乎可以打印出我想要的所有内容以及我想要的方式...（请参阅我的代码）。
我的新代码：
#include <mpi.h>
#include <iostream>
#include <math.h>
#include <string>
#include <cstring>
#include <stdlib.h>
using namespace std;
int main(int argc, char* argv[])
int rang,nbproc, taille;
char name[80];
float time;
double SendData, ReceiveData;
double NumTo6 = 500;
double NumTo7 = 300;
double ReceiveFrom6, ReceiveFrom7;
char message[] = "precu";
int longueur = strlen(message);
int len_buffer = 200;
char Buffer_Time[len_buffer];
char Buffer_Hello[len_buffer];
char Buffer_message[len_buffer];
char receive[] = "none";
int mylen = strlen(receive);
char* Gathered_Char_message = new char[len_buffer];
double DataMessage;
double* GatheredDataMessage = new double[20];
double* GateredDataTime = new double[20];
double DataTime;
int elements[] = ;
int source = 0;
int GatheredSources[] = ;
double NoData = NAN;
SendData = 1254.3356;
ReceiveData = 0;
cout << " " << endl;
MPI_Init(&argc, &argv);
time = MPI_Wtime();
MPI_Comm_size(MPI_COMM_WORLD, &nbproc);
MPI_Comm_rank(MPI_COMM_WORLD, &rang);
MPI_Get_processor_name(name, &taille);
sprintf(Buffer_Hello,"Hello from process %d among %d of the machine %s",rang,nbproc,name);
sprintf(Buffer_Time,"Time elapsed in process %d on %d is " ,rang,nbproc);
sprintf(Buffer_message,"Data received from process ");
MPI_Send(Buffer_Time,len_buffer,MPI_CHAR,0,rang+20,MPI_COMM_WORLD);
MPI_Send(Buffer_Hello,len_buffer,MPI_CHAR,0,rang+10,MPI_COMM_WORLD);
MPI_Send(Buffer_message,len_buffer,MPI_CHAR,0,rang+30,MPI_COMM_WORLD);
if (rang == 1)
DataMessage = 5*6;
source = 1;
if (rang == 2)
MPI_Send(&SendData,1,MPI_DOUBLE,3,1,MPI_COMM_WORLD);
DataMessage = NoData;
if (rang == 3)
MPI_Recv(&ReceiveData,1,MPI_DOUBLE,2,1,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveData;
source = 2;
if (rang == 4)
MPI_Send(&message[1],longueur+1,MPI_CHAR,5,2,MPI_COMM_WORLD);
DataMessage = NoData;
if (rang == 5)
MPI_Recv(&receive,longueur+1,MPI_CHAR,4,2,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = NoData;
source = 4;
// Exchange between 2 processes:
if(rang == 6)
MPI_Sendrecv(&NumTo7,1,MPI_DOUBLE,7,3,&ReceiveFrom7,1,MPI_DOUBLE,7,4,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveFrom7;
elements[rang] = 1;
source = 7;
if(rang == 7)
MPI_Sendrecv(&NumTo6,1,MPI_DOUBLE,6,4,&ReceiveFrom6,1,MPI_DOUBLE,6,3,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveFrom6;
elements[rang] = 1;
source = 6;
DataTime = MPI_Wtime() - time;
MPI_Gather(&DataTime,1,MPI_DOUBLE,GateredDataTime,1,MPI_DOUBLE,0,MPI_COMM_WORLD);
MPI_Gather(&DataMessage,1,MPI_DOUBLE,GatheredDataMessage,1,MPI_DOUBLE,0,MPI_COMM_WORLD);
MPI_Gather(&source,1,MPI_INT,GatheredSources,1,MPI_INT,0,MPI_COMM_WORLD);
// int* recvcounts = new int[nbproc*sizeof(int)];
// MPI_Gather(&mylen,1,MPI_INT,recvcounts,1,MPI_INT,0,MPI_COMM_WORLD);
// int totlen = 0;
// int* displs = new int[nbproc*sizeof(int)];
// //char* totalstring = new char[totlen*sizeof(char)];
// if(rang == 0)
//
// displs[0] = 0;
// totlen += recvcounts[0] + 1;
// for(int i=1; i< nbproc; i++)
//
// totlen += recvcounts[i]+1;
// displs[i] = displs[i-1] + recvcounts[i-1] + 1;
//
//
// char* totalstring = new char[totlen*sizeof(char)];
// if(rang == 0)
//
// for (int i=0; i<totlen-1; i++)
// totalstring[i] = ' ';
// totalstring[totlen-1] = '\0';
//
// MPI_Gatherv(&receive, mylen, MPI_CHAR,
// totalstring, recvcounts, displs, MPI_CHAR,
// 0, MPI_COMM_WORLD);
if(rang == 0)
cout << Buffer_Hello << endl;
for(int i = 1; i < nbproc; i++)
MPI_Recv(Buffer_Hello,len_buffer,MPI_CHAR,i,i+10,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
MPI_Recv(Buffer_message,len_buffer,MPI_CHAR,i,i+30,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << Buffer_Hello << endl;
if(isnan(GatheredDataMessage[i]))
else
cout << Buffer_message << GatheredSources[i] << ": "<<
GatheredDataMessage[i] << endl;
// cout << totalstring[i] << endl;
MPI_Recv(Buffer_Time,len_buffer,MPI_CHAR,i,i+20,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << Buffer_Time << GateredDataTime[i] << " sec" << endl;
cout << " " << endl;
delete[] GatheredDataMessage;
delete[] GateredDataTime;
MPI_Finalize();
return 0;
并输出：
Hello from process 0 among 8 of the machine jeremy-SATELLITE-P50-C
Hello from process 1 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 1: 30
Time elapsed in process 1 on 8 is 0.000248922 sec
Hello from process 2 among 8 of the machine jeremy-SATELLITE-P50-C
Time elapsed in process 2 on 8 is 0.00013139 sec
Hello from process 3 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 2: 1254.34
Time elapsed in process 3 on 8 is 0.000183373 sec
Hello from process 4 among 8 of the machine jeremy-SATELLITE-P50-C
Time elapsed in process 4 on 8 is 0.000121771 sec
Hello from process 5 among 8 of the machine jeremy-SATELLITE-P50-C
Time elapsed in process 5 on 8 is 0.00027475 sec
Hello from process 6 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 7: 500
Time elapsed in process 6 on 8 is 0.00330783 sec
Hello from process 7 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 6: 300
Time elapsed in process 7 on 8 is 0.000215519 sec
所以我接近我想要的，它错过了进程 4 和 5 交换的字符的收集和打印，并为所有其他进程打印出初始字符“无”。例如，我尝试了几件事：https://***.com/a/31932283/14901229， 您可以在我的带有 cmets 的代码中看到这一点，但是没有打印任何内容...
也许你能帮我做最后一件事？另外，如果您看到我的代码的优化方法（我认为有......），请不要犹豫告诉我！
提前致谢！
【问题讨论】：
输出由操作系统处理。您不能从 MPI 中同步它。唯一基于 MPI 的解决方案是将所有文本发送到零处理并让其进行打印。在您的情况下，您还可以MPI_Gather 数字并让进程零打印它们并附带文本。
如果进程4和5有一个独立的打印，那又不能同步。忍受它，或将缓冲区发送到进程零。但实际上，印刷的意义何在？大规模应用程序永远不会精确地并行打印，因为它不能同步，如果它们同步，它们提供了一个明确的机制来收集数据。您还可以使用 shell 机制（您的 mpistarter 是否使用进程号定义环境变量？）将每个进程的输出发送到不同的文件。或者你可以编写一个MPI_printf 函数，作为进程号的前缀。
【参考方案1】：
我终于找到了一种方法来完成所有这些事情，对于那些感兴趣的人，我链接了下面的代码。它可以按我的意愿工作。
但是，我认为这不是更好的方法，所以如果有人可以向我解释如何优化它，我将不胜感激！
这是我的完整代码以及一些解释 cmets：
#include <mpi.h>
#include <iostream>
#include <math.h>
using namespace std;
int main(int argc, char* argv[])
int rank,nbproc, taille;
char name[80];
float time;
double SendData, ReceiveData;
double NumTo6 = 500;
double NumTo7 = 300;
double ReceiveFrom6, ReceiveFrom7;
char message[] = "preceived";
int longueur = strlen(message);
int len_buffer = 200;
char Buffer_Time[len_buffer];
char Buffer_Hello[len_buffer];
char Buffer_message[len_buffer];
char receive[] = "nothing";
char* Gathered_Char_message = new char[len_buffer];
double DataMessage;
double* GatheredDataMessage = new double[20];
double* GateredDataTime = new double[20];
double DataTime;
int source = 0;
int* GatheredSources = new int[20];
double NoData = NAN;
SendData = 1254.3356;
ReceiveData = 0;
cout << " " << endl;
MPI_Init(&argc, &argv);
time = MPI_Wtime();
MPI_Comm_size(MPI_COMM_WORLD, &nbproc);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Get_processor_name(name, &taille);
sprintf(Buffer_Hello,"Hello from process %d among %d of the machine %s",rank,nbproc,name);
sprintf(Buffer_Time,"Time elapsed in process %d on %d is " ,rank,nbproc);
sprintf(Buffer_message,"Data received from process ");
MPI_Send(Buffer_Time,len_buffer,MPI_CHAR,0,rank+20,MPI_COMM_WORLD);
MPI_Send(Buffer_Hello,len_buffer,MPI_CHAR,0,rank+10,MPI_COMM_WORLD);
MPI_Send(Buffer_message,len_buffer,MPI_CHAR,0,rank+30,MPI_COMM_WORLD);
if (rank == 1)
DataMessage = 5*6;
source = 1;
if (rank == 2)
MPI_Send(&SendData,1,MPI_DOUBLE,3,1,MPI_COMM_WORLD);
DataMessage = NoData;
if (rank == 3)
MPI_Recv(&ReceiveData,1,MPI_DOUBLE,2,1,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveData;
source = 2;
if (rank == 4)
MPI_Send(&message[1],longueur+1,MPI_CHAR,5,2,MPI_COMM_WORLD); // takes only a part of the message
DataMessage = NoData;
if (rank == 5)
MPI_Recv(&receive,longueur+1,MPI_CHAR,4,2,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = NoData;
source = 4;
if(rank == 6)
MPI_Sendrecv(&NumTo7,1,MPI_DOUBLE,7,3,&ReceiveFrom7,1,MPI_DOUBLE,7,4,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveFrom7;
source = 7;
if(rank == 7)
MPI_Sendrecv(&NumTo6,1,MPI_DOUBLE,6,4,&ReceiveFrom6,1,MPI_DOUBLE,6,3,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
DataMessage = ReceiveFrom6;
source = 6;
MPI_Gather(&DataMessage,1,MPI_DOUBLE,GatheredDataMessage,1,MPI_DOUBLE,0,MPI_COMM_WORLD);
MPI_Gather(&source,1,MPI_INT,GatheredSources,1,MPI_INT,0,MPI_COMM_WORLD);
int mylen = strlen(receive); // takes back the length of receive
int* recvcounts = new int[nbproc*sizeof(int)]; // stores all the lengths after gathering
MPI_Gather(&mylen,1,MPI_INT,recvcounts,1,MPI_INT,0,MPI_COMM_WORLD); // Gathering all lengths
int totlen = 0; // Total length of the string gathered with spaces
int* displs = new int[nbproc*sizeof(int)]; // stores all spaces
if(rank == 0)
displs[0] = 0; // no space first
totlen += recvcounts[0] + 1; // first word + space
for(int i=1; i < nbproc; i++)
totlen += recvcounts[i]+1; // (+ space and + \0 for the last iteration)
displs[i] = displs[i-1] + recvcounts[i-1] + 1; // previous space + length of previous word + 1 for the \0
char* totalstring = new char[totlen*sizeof(char)]; // Big string with all the 'receive' words and spaces
if(rank == 0)
for (int i=0; i<totlen; i++) // totlen = 65 -> 7*7 + 8 + 8(spaces + \0)
totalstring[i] = ' '; // pre allocation
totalstring[totlen] = '\0'; // end of string
// Gather all the 'receive' words from each process to create a big string with spaces between words
MPI_Gatherv(&receive, mylen, MPI_CHAR,
totalstring, recvcounts, displs, MPI_CHAR,
0, MPI_COMM_WORLD);
char* piece = strtok(totalstring," "); // string function for splitting strings/ char array
//with a delimiter (here a space)
DataTime = MPI_Wtime() - time;
MPI_Gather(&DataTime,1,MPI_DOUBLE,GateredDataTime,1,MPI_DOUBLE,0,MPI_COMM_WORLD);
if(rank == 0)
cout << Buffer_Hello << endl;
cout << "Message received: " << piece << endl;
cout << " " << endl;
for(int i = 1; i < nbproc; i++)
MPI_Recv(Buffer_Hello,len_buffer,MPI_CHAR,i,i+10,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
MPI_Recv(Buffer_message,len_buffer,MPI_CHAR,i,i+30,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << Buffer_Hello << endl;
if(isnan(GatheredDataMessage[i]))
else
cout << Buffer_message << GatheredSources[i] << ": "<<
GatheredDataMessage[i] << endl;
if(piece != NULL) // Stop condition when it reads the \0
piece = strtok(NULL," "); // Alternativelly, a null pointer may be specified,
//in which case the function continues scanning where a previous successful call to the function ended.
cout << "Message received: " << piece << endl;
MPI_Recv(Buffer_Time,len_buffer,MPI_CHAR,i,i+20,MPI_COMM_WORLD,MPI_STATUS_IGNORE);
cout << Buffer_Time << GateredDataTime[i] << " sec" << endl;
cout << " " << endl;
cout << "Time elapsed in process 0 on 8 is " << MPI_Wtime() - time << "sec " << endl;
delete[] GatheredDataMessage;
delete[] GateredDataTime;
MPI_Finalize();
return 0;
这是我的输出：
Hello from process 0 among 8 of the machine jeremy-SATELLITE-P50-C
Message received: nothing
Hello from process 1 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 1: 30
Message received: nothing
Time elapsed in process 1 on 8 is 0.000141763 sec
Hello from process 2 among 8 of the machine jeremy-SATELLITE-P50-C
Message received: nothing
Time elapsed in process 2 on 8 is 0.000560879 sec
Hello from process 3 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 2: 1254.34
Message received: nothing
Time elapsed in process 3 on 8 is 0.0037692 sec
Hello from process 4 among 8 of the machine jeremy-SATELLITE-P50-C
Message received: nothing
Time elapsed in process 4 on 8 is 0.000128876 sec
Hello from process 5 among 8 of the machine jeremy-SATELLITE-P50-C
Message received: received
Time elapsed in process 5 on 8 is 0.00140292 sec
Hello from process 6 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 7: 500
Message received: nothing
Time elapsed in process 6 on 8 is 0.00226591 sec
Hello from process 7 among 8 of the machine jeremy-SATELLITE-P50-C
Data received from process 6: 300
Message received: nothing
Time elapsed in process 7 on 8 is 0.000495762 sec
Time elapsed in process 0 on 8 is 0.00330377sec
【讨论】：
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