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最新推荐文章于 2018-10-26 10:25:51 发布

原创最新推荐文章于 2018-10-26 10:25:51 发布 · 710 阅读

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文章标签：

#thread #iterator #parallel #function

本文介绍了一种简单的函数方法，用于确定程序中适当的线程数量以充分利用多核CPU资源。通过输入迭代次数和单线程最小迭代数，该函数能够计算出最佳线程数量，确保不会超过CPU核心数。

   How to make a program dispatch enough number of threads in order to adapt multiple CPUs? There is a simple function to
solve it about loop logic.

/** get the number of thread that iterator or loop.
    From interator number and CPU number, and one thread's loop number,
    compute out thread number, and make sure that maximum thread number
    is not more than the number of CPUs.

     @param   int n interator / loop number
     @param   int min_n the minimum loop number in one thread
     @return int thread number
*/
int dtn(int n, int min_n)
{
   int max_tn = n / min_n;
   int tn = max_tn > g_ncore ? g_ncore : max_tn; //tn - thread number we need.
   if ( tn < 1 )
   {
        tn = 1;
   }
   return tn;
}

#pragma omp parallel for num_threads(dtn(n, MIN_ITERATOR_NUM))
     for ( i = 0; i < n; i++ )
     {
         printf("Thread Id = %ld/n", omp_get_thread_num());
         //Do some work here
     }