MPI 生成前 20 个数字

Question

这是我尝试生成从 0 开始的前 20 个数字以尝试学习 MPI 的代码。

我的代码如下：

#include <mpi.h>
#include <stdio.h>

int i = 0;
void test(int edge_count){
   while(i < edge_count){
    printf("Edge count %d\n",i);
    i++;
   }
}

int main(int argc, char** argv) {
 int edge_count = 20;
  // int *p = &i;
  // Initialize the MPI environment. The two arguments to MPI Init are not
  // currently used by MPI implementations, but are there in case future
  // implementations might need the arguments.
  MPI_Init(NULL, NULL);

  // Get the number of processes
  int world_size;
  MPI_Comm_size(MPI_COMM_WORLD, &world_size);

  // Get the rank of the process
  int world_rank;
  MPI_Comm_rank(MPI_COMM_WORLD, &world_rank);

  // Get the name of the processor
  char processor_name[MPI_MAX_PROCESSOR_NAME];
  int name_len;
  MPI_Get_processor_name(processor_name, &name_len);

  // Print off a hello world message
  printf("Hello world from processor %s, rank %d out of %d processors\n",
         processor_name, world_rank, world_size);
  test(edge_count);
  printf("The value of i is %d \n",i);

  // Finalize the MPI environment. No more MPI calls can be made after this
  MPI_Finalize();
}

我的输出是：

Hello world from processor ENG401651, rank 0 out of 2 processors
Edge count 0
Edge count 1
Edge count 2
Edge count 3
Edge count 4
Edge count 5
Edge count 6
Edge count 7
Edge count 8
Edge count 9
Edge count 10
Edge count 11
Edge count 12
Edge count 13
Edge count 14
Edge count 15
Edge count 16
Edge count 17
Edge count 18
Edge count 19
The value of i is 20 
Hello world from processor ENG401651, rank 1 out of 2 processors
Edge count 0
Edge count 1
Edge count 2
Edge count 3
Edge count 4
Edge count 5
Edge count 6
Edge count 7
Edge count 8
Edge count 9
Edge count 10
Edge count 11
Edge count 12
Edge count 13
Edge count 14
Edge count 15
Edge count 16
Edge count 17
Edge count 18
Edge count 19
The value of i is 20

我用来运行的代码是：

mpirun -np 2 execFile

我原以为两个处理器都会进行通信并只生成一次从 0 到 19 的数字，但似乎每个处理器都在独立生成自己的一组数字。

我做错了什么？我是 MPI 的新手，无法弄清楚这背后的原因。

Answer 1

计算机只执行您告诉它们的操作。这不仅适用于 MPI，而且适用于任何类型的编程。

您在脚本的哪个位置明确告诉处理器在它们之间分配工作？问题是，你没有。而且它不会自动发生。

您的代码的以下修改版本显示了如何使用 world_size 和 world_rank 让每个进程独立计算它应该执行的工作份额。

为了更好地展示并行性的好处，我使用线程休眠来模拟实际实现中的工作所花费的时间。

#include <mpi.h>
#include <stdio.h>
#include <chrono>
#include <thread>

void test(int start, int end){
  for(int i=start;i<end;i++){
    printf("Edge count %d\n",i);
    //Simulates complicated, time-consuming work
    std::this_thread::sleep_for(std::chrono::milliseconds(500));
  }
}

int main(int argc, char** argv) {
 int edge_count = 20;
  // int *p = &i;
  // Initialize the MPI environment. The two arguments to MPI Init are not
  // currently used by MPI implementations, but are there in case future
  // implementations might need the arguments.
  MPI_Init(NULL, NULL);

  // Get the number of processes
  int world_size;
  MPI_Comm_size(MPI_COMM_WORLD, &world_size);

  // Get the rank of the process
  int world_rank;
  MPI_Comm_rank(MPI_COMM_WORLD, &world_rank);

  // Get the name of the processor
  char processor_name[MPI_MAX_PROCESSOR_NAME];
  int name_len;
  MPI_Get_processor_name(processor_name, &name_len);

  // Print off a hello world message
  printf("Hello world from processor %s, rank %d out of %d processors\n",
         processor_name, world_rank, world_size);

  const int interval   = edge_count/world_size;
  const int iter_start = world_rank*interval;
  const int iter_end   = (world_rank+1)*interval;

  test(iter_start, iter_end);

  // Finalize the MPI environment. No more MPI calls can be made after this
  MPI_Finalize();
}

MPI 生成前 20 个数字

MPI to generate first 20 numbers

c++

mpi