当谈到MPI时,我是一个noob,但这没有任何意义。所以我在这里有一段代码,它使用MPI_Recv和MPI_Send,但是在告诉我我的网格大小之后,在第一个“我在这里创建它”之前,这个东西被冻结了。被发送到屏幕。为什么我的C++ MPI代码冻结在我身上?
我不明白为什么。第一个“我在这里制作”和最后一个输出到屏幕上的东西之间几乎没有任何关系。
这里的代码剪断
void initMesh(double* &phi, double &h, double &riptime,
double &deltat, int &x, int &y, int &xlength, int &ylength, int &tlength, int &ttasks, int &jtasks, int &itasks, int &tstart, int &jstart, int &istart, int &myrank, int &cores) {
int tasksize, remains, tremains;
MPI_Comm_rank(MPI_COMM_WORLD, &myrank);
MPI_Comm_size(MPI_COMM_WORLD, &cores);
if (myrank == 0) {
cout << "How large would you like the mesh"
<<" to be in the x-direction?" << endl;
cin >> x;
cout << "How large would you like the mesh"
<< " to be in the y-direction?\n";
cin >> y;
cout << "What is the distance between each x/y spot h (equal distance)?\n";
cin >> h;
cout << "How much time would you like the program to run for?\n";
cin >> riptime;
cout << "What would you like the time-step for the analysis to be?\n";
cin >> deltat;
xlength = (int) (x/h);
ylength = (int) (y/h);
tlength = (int) (riptime/deltat);
cout << "Mesh x-points = " << xlength << endl;
cout << "Mesh y-points = " << ylength << endl;
cout << "Mesh time points = " << tlength << endl;
cout << "I made it here!";
}
//GOOD UP TO HERE!!! Then it freezes when I run the thing with 3 or more processors
for (int i=1; i < cores; i++) {
if (myrank==0) {
cout << "I made it here!";
MPI_Send(&xlength, 1, MPI_INT, i, 0, MPI_COMM_WORLD);
}
else {
MPI_Recv(&xlength, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
cout << "I made it here!";
}
}
for (int i=1; i < cores; i++) {
if (myrank==0) {
MPI_Send(&ylength, 1, MPI_INT, i, 1, MPI_COMM_WORLD);
}
else {
MPI_Recv(&ylength, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
}
}
for (int i=1; i < cores; i++) {
if (myrank==0) {
MPI_Send(&tlength, 1, MPI_INT, i, 2, MPI_COMM_WORLD);
}
else {
MPI_Recv(&tlength, 1, MPI_INT, 0, 2, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
}
}
cout << "I made it here!";
代码的上述部分就是麻烦的是现在。
第一个''我把它放在这里!'''被缓冲(没有'std :: endl'来刷新)。如果你解决了这个问题会怎样? – chrisaycock 2013-04-28 22:10:23
对不起,我是这个东西的新手。你是什么意思由std :: endl刷新?我注意到,当我用(int)部分的时候,我说隐藏了其余的代码后,它被零除。 – Mechy 2013-04-28 22:11:23
尝试使用'cerr'而不是所有'cout',并在行尾加上'<< std :: endl;'。这将确保您的输出不被缓冲,而是直接写入屏幕。 @chrisaycock怀疑,最后一行没有打印的事实只是缓冲问题,并不是因为程序在产生输出的行之前失败。 – 2013-04-28 22:14:43