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是否可以std::inner_product()从 C++ 与omp.h库并行化？不幸的是，我无法__gnu_parallel::inner_product()在较新版本的 gcc 中使用 available 。我知道我可以实现自己的inner_product并将其并行化，但我想使用标准方法。

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2 回答 2

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简短的回答：没有。

像这样的算法的全部意义inner_product在于它们将循环从你身上抽象出来。但为了并行化算法，您需要并行化该循环——通过#pragma omp parallel for或通过并行部分。这两种方法在本质上都与代码结构中的循环相关联，因此即使循环可并行化（很可能），您也需要将 OpenMP pragma放入函数中以对其应用并行性。

于 2012-12-07T13:29:57.617 回答
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跟进 Hristo 的评论，您可以通过在线程inner_product上分解数组，调用每个子数组，然后使用某种归约操作来组合子结果来做到这一点

#include <iostream>
#include <numeric>
#include <omp.h>

#include <sys/time.h>
void tick(struct timeval *t);
double tock(struct timeval *t);

int main (int argc, char **argv) {
  const long int nelements=1000000;
  long int *a = new long int[nelements];
  long int *b = new long int[nelements];
  int nthreads;
  long int sum = 0;
  struct timeval t;
  double time;

  #pragma omp parallel for
  for (long int i=0; i<nelements; i++) {
        a[i] = i+1;
        b[i] = 1;
  }

  tick(&t);
  #pragma omp parallel 
  #pragma omp single
  nthreads = omp_get_num_threads();

  #pragma omp parallel default(none) reduction(+:sum) shared(a,b,nthreads) 
  {
       int tid = omp_get_thread_num();
       int nitems = nelements/nthreads;
       int start = tid*nitems;
       int end   = start + nitems;
       if (tid == nthreads-1) end = nelements;

       sum += std::inner_product( &(a[start]), a+end, &(b[start]), 0L);
  }
  time = tock(&t);

  std::cout << "using omp: sum = " << sum << " time = " << time << std::endl;

  delete [] a;
  delete [] b;



  a = new long int[nelements];
  b = new long int[nelements];
  sum = 0;

  for (long int i=0; i<nelements; i++) {
        a[i] = i+1;
        b[i] = 1;
  }
  tick(&t);
  sum = std::inner_product( a, a+nelements, b, 0L);
  time = tock(&t);

  std::cout << "single threaded: sum = " << sum << " time = " << time << std::endl;

  std::cout << "correct answer: sum = " << (nelements)*(nelements+1)/2 << std::endl ;

  delete [] a;
  delete [] b;

  return 0;
}

void tick(struct timeval *t) {
    gettimeofday(t, NULL);
}

/* returns time in seconds from now to time described by t */
double tock(struct timeval *t) {
    struct timeval now;
    gettimeofday(&now, NULL);
    return (double)(now.tv_sec - t->tv_sec) + ((double)(now.tv_usec - t->tv_usec)/1000000.);
}

运行它得到的加速比我预期的要好：

$ for NT in 1 2 4 8; do export OMP_NUM_THREADS=${NT}; echo; echo "NTHREADS=${NT}";./inner; done

NTHREADS=1
using omp: sum = 500000500000 time = 0.004675
single threaded: sum = 500000500000 time = 0.004765
correct answer: sum = 500000500000

NTHREADS=2
using omp: sum = 500000500000 time = 0.002317
single threaded: sum = 500000500000 time = 0.004773
correct answer: sum = 500000500000

NTHREADS=4
using omp: sum = 500000500000 time = 0.001205
single threaded: sum = 500000500000 time = 0.004758
correct answer: sum = 500000500000

NTHREADS=8
using omp: sum = 500000500000 time = 0.000617
single threaded: sum = 500000500000 time = 0.004784
correct answer: sum = 500000500000
于 2012-12-07T14:48:18.253 回答