問題描述

我正在嘗試并行化以下程序，但不知道如何減少數(shù)組.我知道這是不可能的，但有沒有其他選擇?謝謝.(我在 m 上添加了reduce，這是錯誤的，但想就如何做到這一點(diǎn)提出建議.)

I am trying to parallelize the following program, but don't know how to reduce on an array. I know it is not possible to do so, but is there an alternative? Thanks. (I added reduction on m which is wrong but would like to have an advice on how to do it.)

#include <iostream>
#include <stdio.h>
#include <time.h>
#include <omp.h>
using namespace std;

int main ()
{
  int A [] = {84, 30, 95, 94, 36, 73, 52, 23, 2, 13};
  int S [10];

  time_t start_time = time(NULL);
  #pragma omp parallel for private(m) reduction(+:m)
  for (int n=0 ; n<10 ; ++n ){
    for (int m=0; m<=n; ++m){
      S[n] += A[m];
    }
  }
  time_t end_time = time(NULL);
  cout << end_time-start_time;

  return 0;
}

推薦答案

是的，可以使用 OpenMP 進(jìn)行數(shù)組縮減.在 Fortran 中，它甚至為此有構(gòu)造.在 C/C++ 中，你必須自己做.這里有兩種方法可以做到.

Yes it is possible to do an array reduction with OpenMP. In Fortran it even has construct for this. In C/C++ you have to do it yourself. Here are two ways to do it.

第一種方法為每個線程制作私有版本的S，并行填充，然后在臨界區(qū)合并成S(見下面的代碼).第二種方法創(chuàng)建一個維度為 10*nthreads 的數(shù)組.并行填充此數(shù)組，然后將其合并到 S 中，而不使用臨界區(qū).第二種方法要復(fù)雜得多，如果您不小心，可能會出現(xiàn)緩存問題，尤其是在多路系統(tǒng)上.有關(guān)更多詳細(xì)信息，請參閱此填充直方圖(數(shù)組縮減)與 OpenMP 并行，無需使用臨界區(qū)

The first method makes private version of S for each thread, fill them in parallel, and then merges them into S in a critical section (see the code below). The second method makes an array with dimentions 10*nthreads. Fills this array in parallel and then merges it into S without using a critical section. The second method is much more complicated and can have cache issues especially on multi-socket systems if you are not careful. For more details see this Fill histograms (array reduction) in parallel with OpenMP without using a critical section

第一種方法

int A [] = {84, 30, 95, 94, 36, 73, 52, 23, 2, 13};
int S [10] = {0};
#pragma omp parallel
{
    int S_private[10] = {0};
    #pragma omp for
    for (int n=0 ; n<10 ; ++n ) {
        for (int m=0; m<=n; ++m){
            S_private[n] += A[m];
        }
    }
    #pragma omp critical
    {
        for(int n=0; n<10; ++n) {
            S[n] += S_private[n];
        }
    }
}

第二種方法

int A [] = {84, 30, 95, 94, 36, 73, 52, 23, 2, 13};
int S [10] = {0};
int *S_private;
#pragma omp parallel
{
    const int nthreads = omp_get_num_threads();
    const int ithread = omp_get_thread_num();

    #pragma omp single 
    {
        S_private = new int[10*nthreads];
        for(int i=0; i<(10*nthreads); i++) S_private[i] = 0;
    }
    #pragma omp for
    for (int n=0 ; n<10 ; ++n )
    {
        for (int m=0; m<=n; ++m){
            S_private[ithread*10+n] += A[m];
        }
    }
    #pragma omp for
    for(int i=0; i<10; i++) {
        for(int t=0; t<nthreads; t++) {
            S[i] += S_private[10*t + i];
        }
    }
}
delete[] S_private;

這篇關(guān)于減少 OpenMP 中的數(shù)組的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！