我想寻求一些帮助,以完成mpi实现中的工作量分配。
想法是节点0将具有文件名列表(“ list_file”)。当其他节点空闲时,它们将向节点0发送文件请求,而节点0将发回文件名。
一旦没有更多文件可发送,节点0完成其工作。但是,如何向其他节点的读取器线程发送信号,通知节点0没有更多文件,它们应该停止等待节点0发送新文件。
// pid 0 will collect all input text file names and distribute them to other nodes
if (pid == 0)
{
vector<char*> list_file; // a vector to hold the input text file names to be processed
GetInputFile(argv, list_file); // argv[1] is a text file that contains the input text files to be processed, all the text file names will be added to list_file
num_file_remaining = list_file.size(); // number of file remained in the queue to be processed
MPI_Status requeststats; // for MPI_recv
// listen to request for file from other nodes as long as there is file left in list_file
while (num_file_remaining != 0)
{
MPI_recv(NULL, 0, MPI_INT, MPI_ANY_SOURCE, 0, MPI_COMM_WORLD, &requeststats); // listen to request for file
MPI_send(list_file.back(), 5 * MAX_WORD_LENGTH, MPI_CHAR, requeststats.MPI_SOURCE, requeststats.MPI_SOURCE, MPI_COMM_WORLD); // send the file to respective node
list_file.pop_back(); // remove the file that was just sent
num_file_remaining -= 1; // reduce the number of file remained in the queue
}
}
// other nodes will request work from pid 0
if (pid != 0)
{
char* file_name;
while (num_file_remaining != 0)
{
MPI_send(NULL, 0, MPI_INT, 0, 0, MPI_COMM_WORLD); // send the request for a file to node 0
MPI_recv(file_name, 5 * MAX_WORD_LENGTH, MPI_CHAR, 0, pid, MPI_COMM_WORLD, MPI_STATUS_IGNORE); // receive the file from node 0
cout << "pid: " << pid << " - " << file_name << endl; // process the file
// HOW TO EXIT THE LOOP WHEN NO MORE FILE TO RECEIVE FROM NODE 0
}
}
答案 0 :(得分:0)
有一个有效的MPI_Irecv
通话:
// set this tag to whatever you want and don’t use it for anything else
constexpr int AM_I_FINISHED_TAG = 1376;
MPI_Request am_I_finished = MPI_Irecv(nullptr, 0, MPI_INT, 0, AM_I_FINISHED_TAG, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
每当在子节点上完成文件处理后,请使用MPI_Test检查该请求是否完成。当主节点不再有子节点的工作时,请使用AM_I_FINISHED_TAG将数据发送到所有子节点。