C语言 为什么这个 MPI 程序停滞不前

Question

//#define SIZE 3
void master(int n,int size)
{
for(int j=1;j<size;j++){
    MPI_Send(&n,1,MPI_INT,j,1,MPI_COMM_WORLD);
    printf("n Sent %d to process %d",n,j);
    fflush(stdout);
    }
}
void slave(int size)
{
for(int j=1;j<size;j++){
    int k=0;
    MPI_Status status;
    MPI_Recv(&k,1,MPI_INT,0,1,MPI_COMM_WORLD,&status);
    printf("n Process %d has received %d",j,k);
    fflush(stdout);
    }
}
int main(int argc,char** argv)
{
MPI_Init(&argc,&argv);
int la_size;
int rank;
MPI_Comm_size(MPI_COMM_WORLD,&la_size);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);
for(int i=0;i<3;i++){
    if(rank==0)
        master(i,la_size);
    else
        slave(la_size);
}
MPI_Finalize();
printf("nprogram finished...");
fflush(stdout);
return 0;
}  

上面的程序看起来很简单，但它停滞不前。是僵局吗？输出为：

     Sent 0 to process 1  
 Sent 0 to process 2  
 Sent 1 to process 1  
 Sent 1 to process 2  
 Process 1 has received 0  
 Process 2 has received 1  
 Process 1 has received 2  
 Sent 2 to process 1  
 Sent 2 to process 2  
 Process 1 has received 0  
 Process 2 has received 1  
 Process 1 has received 2

Answer 1

在循环的每次迭代中 main 中，等级 0 对许多从属等级中的每一个执行一次发送，但每个从属站发布的接收数量与总共从属等级的数量一样多。由于没有已发布的发送来匹配后面的接收，因此接收方无限期地阻塞，程序挂起。

Answer 2

多亏了Novelocrat的回答，正确的代码实际上是这样的：

#include <stdio.h>
#include <mpi.h>
#include <stdlib.h>
//#define SIZE 3
void master(int n,int size)
{
    for(int j=1;j<size;j++){
    MPI_Send(&n,1,MPI_INT,j,1,MPI_COMM_WORLD);
    printf("n Sent %d to process %d",n,j);
    fflush(stdout);
    }
}
void slave(int size)
{
    int k=0,rank=0;
    MPI_Comm_rank(MPI_COMM_WORLD,&rank);
    MPI_Status status;
    MPI_Recv(&k,1,MPI_INT,0,1,MPI_COMM_WORLD,&status);
    printf("n Process %d has received %d",rank,k);
    fflush(stdout);
}
int main(int argc,char** argv)
{
    MPI_Init(&argc,&argv);
    int la_size;
    int rank;
    MPI_Comm_size(MPI_COMM_WORLD,&la_size);
    MPI_Comm_rank(MPI_COMM_WORLD,&rank);
    for(int i=0;i<3;i++){
    if(rank==0)
        master(i,la_size);
    else
        slave(la_size);
    }
    MPI_Finalize();
    printf("nprogram finished...");
    fflush(stdout);
    return 0;
}