我正在尝试学习如何使用OpenMPI,并遇到了此示例代码
#include "mpi.h”
int main(int argc, char **argv)
{
// Call MPI initialization
MPI_Init(&argc, &argv);
// Get my processor ID and number of processor
int myProcID;
int numProcs;
MPI_Comm_rank(MPI_COMM_WORLD, &myProcID);
MPI_Comm_size(MPI_COMM_WORLD, &numProcs);
// Set number of cells from command line argument
int numCells = atoi(argv[1]);
<etc…>
// Find which piece of the integral this
// processor is responsible for
int numCellsPerProc = numCells/numProcs;
int myStart = myProcID * numCellsPerProc;
int myEnd = myStart + numCellsPerProc;
// Account for unequal load by making sure
// last processor has correct end
if (myProcID == numProcs - 1) myEnd = numCells;
// Loop over cells and compute integral
// using trapezoidal rule
double myResult = 0.0;
for (int i = myStart; i < myEnd; ++i)
{
double xL = xMin + i*dx;
double xR = xL + dx;
myResult += 0.5 * (myFunction(xL)+myFunction(xR)) * dx;
}
// Sum result across processors
double totalResult;
MPI_Reduce(&myResult, &totalResult, 1, MPI_DOUBLE, MPI_SUM,
0, MPI_COMM_WORLD);
// Print the result, but only from root processor
if (myProcID == 0)
{
std::cout << "result = ";
std::cout << std::fixed << std::setprecision(10)
<< totalResult << std::endl;
}
// Call MPI_Finalize
MPI_Finalize();
return 0;
}
<etc>
原谅我对处理器实际架构的无知。为什么示例代码设置了单元数?我以为每个处理器总体上一次负责一项工作?
我完全不明白这些说法...
// Set number of cells from command line argument
int numCells = atoi(argv[1]);
<etc…>
// Find which piece of the integral this
// processor is responsible for
int numCellsPerProc = numCells/numProcs;
int myStart = myProcID * numCellsPerProc;
int myEnd = myStart + numCellsPerProc
最佳答案
这取决于命令行参数-argv[1]
-每个节点将有多少个作业(例如,在OpenMPI中,您可以通过-N
指定每个节点的作业数)。此外,您可以生成线程以利用多核处理器。
实际上,您是在计算积分。您将积分间隔分为numProcs
个部分,因此每个作业都会计算出自己的部分,最后,所有结果都由缩减项求和。
(单词cell
-在这种情况下不是很好的变量名)
关于c++ - 为什么在尝试实现OpenMPI时设置单元数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46351909/