(c)MPI流程在循环后无法做出响应
假设我有一个代码块,该代码块试图计算斐波那契序列中的前15个数字,并使用for循环在3个过程(mpi_send)之间分配每个唯一数字,如下所示。
int main(int argc, char* argv[]) {
int rank, size, recieve_data1, recieve_data2;
MPI_Init(NULL, NULL);
MPI_Status status;
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &size);
printf("Available ranks are: %d \n \n", rank); // first rank rollcall
fflush(stdout);
int num1 = 1; int num2 = 1;
int RecieveNum; int SumNum;
for
(int n = 0; n < 16; n++) {
if
(rank == 0) {
// perform the fibb sequence algorithim
SumNum = num1 + num2;
num1 = num2;
num2 = SumNum;
// define the sorting algorithim
int DeliverTo = (n % 3) + 1;
// send calculated result
MPI_Send(&SumNum, 1, MPI_INT, DeliverTo, 1, MPI_COMM_WORLD);
}
else {
// recieve the element integer
MPI_Recv(&RecieveNum, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
// print and flush the buffer
printf("I am process rank %d, and I recieved the number %d. \n", rank, RecieveNum);
fflush(stdout);
}
}
printf("Available ranks are: %d \n \n", rank); // second rank roll call
fflush(stdout);
/*
more code that I run...
*/
MPI_Finalize();
return 0;
}
在调用第一个循环之前,进程0,1,2,3所有响应printf(可用等级为:%d \ n \ n“,等级);
在第25行中。 ,在执行第一个循环并使用第二个printf
之后,只有进程0响应我,我期望所有4-3个进程在执行for for循环后再次响应。为了解决这个问题,我隔离了这一部分,并试图在没有成功的情况下进行调试。 的数字。
访问此序列生成 出现错误,从我看到的东西(如果我错了,请纠正我),所有过程都在完成for循环后悬挂,但是我不确定为什么或如何修复
网站 。这个问题(从我的角度来看)。我有点像MPI(和堆栈溢出)新手,因此也欢迎任何代码指针。谢谢
Suppose that I have a code block that attempts to calculate the first 15 numbers in the Fibonacci sequence and distributes each unique number among 3 processes (MPI_Send) using a for loop as shown in the code block below.
int main(int argc, char* argv[]) {
int rank, size, recieve_data1, recieve_data2;
MPI_Init(NULL, NULL);
MPI_Status status;
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &size);
printf("Available ranks are: %d \n \n", rank); // first rank rollcall
fflush(stdout);
int num1 = 1; int num2 = 1;
int RecieveNum; int SumNum;
for
(int n = 0; n < 16; n++) {
if
(rank == 0) {
// perform the fibb sequence algorithim
SumNum = num1 + num2;
num1 = num2;
num2 = SumNum;
// define the sorting algorithim
int DeliverTo = (n % 3) + 1;
// send calculated result
MPI_Send(&SumNum, 1, MPI_INT, DeliverTo, 1, MPI_COMM_WORLD);
}
else {
// recieve the element integer
MPI_Recv(&RecieveNum, 1, MPI_INT, 0, 1, MPI_COMM_WORLD, &status);
// print and flush the buffer
printf("I am process rank %d, and I recieved the number %d. \n", rank, RecieveNum);
fflush(stdout);
}
}
printf("Available ranks are: %d \n \n", rank); // second rank roll call
fflush(stdout);
/*
more code that I run...
*/
MPI_Finalize();
return 0;
}
Before the first for loop is called, processes 0,1,2,3 all respond to the printf(Available ranks are: %d \n \n", rank);
on line 25. However, after executing the first for loop and using a second printf
, only process 0 responds. I was expecting all 4 processes 0 - 3 to respond again after the execution of the for loop. To solve this problem, I isolated this section of code and attempted to debug for several hours with no sucess. This particular issue proves to be problematic, as I have additional code (not shown here for the sake of being concise), that will access the numbers generated from this sequence.
Finally, I am running the code by building the solution, running the VS terminal as an administrator, and typing mpiexec -n 4 my_file_name.exe
. No build errors or complication mistakes occurred. From what I can see (correct me if I'm wrong), all processes hang after completing the for loop, but I am unsure why or how to fix it.
After searching the website, I did not see anything that answered this question (from my point of view). I am a bit of an MPI (and Stack Overflow) newbie, so any code pointers are also welcomed. Thanks
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您的流程零计算要发送给谁。然后每个人都接受。这意味着所有不是计算的接收器的过程都将悬挂。
在MPI中,将您发送到动态计算的接收器的情况并不容易。您要么需要
mpi_put的单方面操作
计算机接收器上的数据。You have process zero compute who to send to. And then everyone does a receive. That means that all processes that are not the computed receiver will hang.
This scenario where you send to a dynamically computed receiver is not easy to do in MPI. You either need to
MPI_Put
the data on the computer receiver.