如何使用MPI计算数组的总和（代码实现） _MPI_Recv

先决条件： MPI –简化分布式计算
消息传递接口(MPI)
是例程的库, 可用于在C或Fortran77中创建并行程序。它允许用户通过创建并行进程来构建并行应用程序, 并在这些进程之间交换信息。
MPI使用两个基本的通信例程：

MPI_Send, 将消息发送到另一个进程。
MPI_Recv, 以接收来自另一个进程的消息。

MPI_Send和MPI_Recv的语法为：

int MPI_Send(void *data_to_send, int send_count, MPI_Datatype send_type, int destination_ID, int tag, MPI_Comm comm); int MPI_Recv(void *received_data, int receive_count, MPI_Datatype receive_type, int sender_ID, int tag, MPI_Comm comm, MPI_Status *status);

为了降低程序的时间复杂度, 子数组的并行执行是通过运行并行进程来计算其部分和, 然后, 主进程(根进程)计算这些部分和的总和, 以返回子数组的总和。数组。
【如何使用MPI计算数组的总和（代码实现）】例子：

Input : {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Output : Sum of array is 55Input : {1, 3, 5, 10, 12, 20, 4, 50, 100, 1000} Output : Sum of array is 1205

注意 -你必须在基于Linux的系统上安装MPI才能执行以下程序。有关详细信息, 请参阅MPI –简化分布式计算
使用以下代码编译并运行程序：

mpicc program_name.c -o object_file mpirun -np [number of processes] ./object_file

以下是上述主题的实现：

#include < mpi.h> #include < stdio.h> #include < stdlib.h> #include < unistd.h> // size of array #define n 10int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; // Temporary array for slave process int a2[1000]; int main( int argc, char * argv[]) {int pid, np, elements_per_process, n_elements_recieved; // np -> no. of processes // pid -> process idMPI_Status status; // Creation of parallel processes MPI_Init(& argc, & argv); // find out process ID, // and how many processes were started MPI_Comm_rank(MPI_COMM_WORLD, & pid); MPI_Comm_size(MPI_COMM_WORLD, & np); // master process if (pid == 0) { int index, i; elements_per_process = n / np; // check if more than 1 processes are run if (np > 1) { // distributes the portion of array // to child processes to calculate // their partial sums for (i = 1; i < np - 1; i++) { index = i * elements_per_process; MPI_Send(& elements_per_process, 1, MPI_INT, i, 0, MPI_COMM_WORLD); MPI_Send(& a[index], elements_per_process, MPI_INT, i, 0, MPI_COMM_WORLD); }// last process adds remaining elements index = i * elements_per_process; int elements_left = n - index; MPI_Send(& elements_left, 1, MPI_INT, i, 0, MPI_COMM_WORLD); MPI_Send(& a[index], elements_left, MPI_INT, i, 0, MPI_COMM_WORLD); }// master process add its own sub array int sum = 0; for (i = 0; i < elements_per_process; i++) sum += a[i]; // collects partial sums from other processes int tmp; for (i = 1; i < np; i++) { MPI_Recv(& tmp, 1, MPI_INT, MPI_ANY_SOURCE, 0, MPI_COMM_WORLD, & status); int sender = status.MPI_SOURCE; sum += tmp; }// prints the final sum of array printf ( "Sum of array is : %d\n" , sum); } // slave processes else { MPI_Recv(& n_elements_recieved, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, & status); // stores the received array segment // in local array a2 MPI_Recv(& a2, n_elements_recieved, MPI_INT, 0, 0, MPI_COMM_WORLD, & status); // calculates its partial sum int partial_sum = 0; for ( int i = 0; i < n_elements_recieved; i++) partial_sum += a2[i]; // sends the partial sum to the root process MPI_Send(& partial_sum, 1, MPI_INT, 0, 0, MPI_COMM_WORLD); }// cleans up all MPI state before exit of process MPI_Finalize(); return 0; }

输出如下：