Question

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

int main(argc,argv)
int argc;
char *argv[];  {
int numtasks, rank, dest, source, rc, count, tag=1;
MPI_Status Stat;

MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD, &numtasks);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);

if (rank%2 == 1) {
dest = (rank+1)%numtasks;
  source = (rank-1+numtasks)%numtasks;
  rc = MPI_Send(&outmsg, 1, MPI_CHAR, dest, tag, MPI_COMM_WORLD);
  rc = MPI_Recv(&inmsg, 1, MPI_CHAR, source, tag,MPICOMM_WORLD, &Stat);
}

else  {
  dest   = (rank-1+numtasks)%numtasks;
  source = (rank+1)%numtasks;
  rc = MPI_Recv(  &inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat );
  rc = MPI_Send( &outmsg, 1, MPI_CHAR, dest,   tag, MPI_COMM_WORLD );
}
rc = MPI_Get_count(&Stat, MPI_CHAR, &count);
printf("Task %d: Received %d char(s) from task %d with tag %d\n",
   rank, count, Stat.MPI_SOURCE, Stat.MPI_TAG);

MPI_Finalize();
}

结果：

Task 2: Received 1 char(s) from task 3 with a tag 1
Task 0: Received 1 char(s) from task 1 with a tag 1

为什么任务无法识别2个以上的处理器？我想在超过2个处理器上运行它我用环模式更新了mpi程序，我认为如果（排名％2 == 1）

，行可能出错了

Answer 1

count和Stat仅在0级和1级初始化，因此您基本上在第2和第3级上打印垃圾。

你的程序是硬编码的，可以在2个任务上运行（它会在1个任务上崩溃，并且打印3个或更多任务的垃圾）。

我建议您使用MPI_Sendrecv()进行调查，以便在一次MPI通话中发送和接收无死锁。

不确定您要实现的目标，请详细说明并更新您的问题。话虽这么说，您可以考虑使用响铃模式，以便您的程序可以在任意数量的MPI任务上运行（例如发送到(rank+1)%numtasks并从(rank+numtasks-1)%numtasks接收）

根据您编辑的问题，所有MPI等级的通信模式可以如下所示

dest = (rank+1)%numtasks;
source = (rank-1+numtasks)%numtasks;
MPI_Sendrecv(&outmsg, 1, MPI_CHAR, dest, tag,
             &inmsg, 1, MPI_CHAR, source, tag,
             MPI_COMM_WORLD, &Stat);

Answer 2

可能会尝试更好地调查MPI机制：

不要犹豫尝试代码，因为它会帮助您掌握 numtasks ， rank -ID的概念并将其关联起来运行时操作。编译错误将在上面发布的代码中报告快捷方式。要获得更多与MPI相关的做法和见解，还应从相应的rc调用

分配和评估 MPI_<fun>() 返回代码

问：为什么任务无法识别2个以上的处理器？答： 可以，，但if(){..}else if(){..}代码块不允许其他人产生任何可见的输出。

问：我想将4197005的结果更改为2,3 - 如何？
A：有人无法更改MPI报告的数字，但您可以更改代码的行为（请参见下文），并在可行的情况下进行更多输出谁正在执行什么，时， 实际上是什么。这样我们就可以学习理解MPI概念并验证实际的MPI代码执行。

随意提出更多问题。

#include "mpi.h" #include <stdio.h> int main( int argc, char *argv[] ) { int numtasks, rank, dest, source, rc, count, tag = 1; MPI_Status Stat; printf( "(INF) will MPI_Init()...\n" ); MPI_Init( &argc, &argv ); printf( "(INF) MPI_Init() done.\n" ); MPI_Comm_size( MPI_COMM_WORLD, &numtasks ); MPI_Comm_rank( MPI_COMM_WORLD, &rank ); printf( "(INF) I am MPI[%d of %d]\n", rank, numtasks ); if ( rank == 0 ) { dest = 1; source = 1; rc = MPI_Send( &outmsg, 1, MPI_CHAR, dest, tag, MPI_COMM_WORLD ); rc = MPI_Recv( &inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat ); printf( "(INF) I am MPI[%d of %d] having just completed an MPI_Send( ->MPI[%d] ) + MPI_Recv( <-MPI[%d] ). Last RetCODE == %d.\n", rank, numtasks, dest, source, rc ); } else if ( rank == 1 ) { dest = 0; source = 0; rc = MPI_Recv( &inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat ); rc = MPI_Send( &outmsg, 1, MPI_CHAR, dest, tag, MPI_COMM_WORLD ); printf( "(INF) I am MPI[%d of %d] having just completed an MPI_Recv( <-MPI[%d] ) + MPI_Send( ->MPI[%d] ). Last RetCODE == %d\n", rank, numtasks, source, dest, rc ); } else if ( rank == 2 ) { dest = 3; source = 3; rc = MPI_Recv( &inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat ); rc = MPI_Send( &outmsg, 1, MPI_CHAR, dest, tag, MPI_COMM_WORLD ); printf( "(INF) I am MPI[%d of %d] having just completed an MPI_Recv( <-MPI[%d] ) + MPI_Send( ->MPI[%d] ). Last RetCODE == %d\n", rank, numtasks, source, dest, rc ); } else if ( rank == 3 ) { dest = 2; source = 2; rc = MPI_Send( &outmsg, 1, MPI_CHAR, dest, tag, MPI_COMM_WORLD ); rc = MPI_Recv( &inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat ); printf( "(INF) I am MPI[%d of %d] having just completed an MPI_Send( ->MPI[%d] ) + MPI_Recv( <-MPI[%d] ). Last RetCODE == %d\n", rank, numtasks, dest, source, rc ); } else { printf( "(INF) I am MPI[%d of %d] will NOP...\n", rank, numtasks ); } rc = MPI_Get_count( &Stat, MPI_CHAR, &count ); /* Task 3: Received 1 char(s) from task 4197005 with a tag 0 Task 2: Received 1 char(s) from task 4197005 with a tag 0 Task 1: Received 1 char(s) from task 0 with a tag 1 Task 0: Received 1 char(s) from task 1 with a tag 1 */ printf( "Task %d: Received %d char(s) from task %d with tag %d\n", rank, count, Stat.MPI_SOURCE, Stat.MPI_TAG ); printf( "(INF) I am MPI[%d of %d]: will MPI_Finalize()...\n", rank, numtasks ); MPI_Finalize(); printf( "(INF) I am MPI[%d of %d]: MPI_Finalize() done, will exit()...\n", rank, numtasks ); }

Answer 3

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

int main (argc,argv)
int argc;
char *argv[]; {
int numtasks, rank, dest, source, rc, count, tag=1;
char inmsg, outmsg='x';
MPI_Status Stat;

MPI_Init (&argc, &argv);
MPI_Comm_size (MPI_COMM_WORLD, &numtasks);
MPI_Comm_rank (MPI_COMM_WORLD, &rank);

if (rank%2 == 0)
{
dest=(rank+1)%numtasks;
source=(rank+1)%numtasks;

rc = MPI_Send(&outmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD);
rc = MPI_Recv(&inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat);
}

else {
dest = (rank-1+numtasks)%numtasks;
source = (rank-1+numtasks)%numtasks;
rc = MPI_Recv (&inmsg, 1, MPI_CHAR, source, tag, MPI_COMM_WORLD, &Stat);
rc = MPI_Send (&outmsg, 1, MPI_CHAR, dest,  tag, MPI_COMM_WORLD);
}

{
rc = MPI_Get_count (&Stat, MPI_CHAR, &count);
printf ("Task %d: Received %d char(s) from task %d with a tag %d \n", rank, count, Stat.MPI_SOURCE, Stat.MPI_TAG);
}
MPI_Finalize();

MPI发送接收

3 个答案:

可能会尝试更好地调查MPI机制：