我有以下称为 pgain 的方法，它调用我试图并行化的方法 dist:

/******************************************************************************/

/* For a given point x, find the cost of the following operation:
 * -- open a facility at x if there isn't already one there,
 * -- for points y such that the assignment distance of y exceeds dist(y, x),
 * make y a member of x,
 * -- for facilities y such that reassigning y and all its members to x
 * would save cost, realize this closing and reassignment.
 *
 * If the cost of this operation is negative (i.e., if this entire operation
 * saves cost), perform this operation and return the amount of cost saved;
 * otherwise, do nothing.
 */

/* numcenters will be updated to reflect the new number of centers */
/* z is the facility cost, x is the number of this point in the array
   points */
double pgain ( long x, Points *points, double z, long int *numcenters )
{
    số nguyên i;
    int number_of_centers_to_close = 0;

    static double *work_mem;
    static double gl_cost_of_opening_x;
    static int gl_number_of_centers_to_close;

    int stride = *numcenters + 2;
    //make stride a multiple of CACHE_LINE
    int cl = CACHE_LINE/sizeof ( double );
    if ( stride % cl != 0 ) {
        stride = cl * ( stride / cl + 1 );
    }
    int K = stride - 2 ; // K==*numcenters

    //my own cost of opening x
    double cost_of_opening_x = 0;

    work_mem = ( double* ) malloc ( 2 * stride * sizeof ( double ) );
    gl_cost_of_opening_x = 0;
    gl_number_of_centers_to_close = 0;

    /*
     * For each center, we have a *lower* field that indicates
     * how much we will save by closing the center.
     */
    int đếm = 0;
    for ( int i = 0; i < points->num; i++ ) {
        if ( is_center[i] ) {
            center_table[i] = count++;
        }
    }
    work_mem[0] = 0;

    //now we finish building the table. clear the working memory.
    memset ( switch_membership, 0, points->num * sizeof ( bool ) );
    memset ( work_mem, 0, stride*sizeof ( double ) );
    memset ( work_mem+stride,0,stride*sizeof ( double ) );

    //my *lower* fields
    double* lower = &work_mem[0];
    //global *lower* fields
    double* gl_lower = &work_mem[stride];

    #pragma omp parallel for
    for ( i = 0; i < points->num; i++ ) {
        float x_cost = dist ( points->p[i], points->p[x], points->dim ) * points->p[i].weight;
        float current_cost = points->p[i].cost;

        if ( x_cost < current_cost ) {

            // point i would save cost just by switching to x
            // (note that i cannot be a median,
            // or else dist(p[i], p[x]) would be 0)

            switch_membership[i] = 1;
            cost_of_opening_x += x_cost - current_cost;

        } khác {

            // cost of assigning i to x is at least current assignment cost of i

            // consider the savings that i's **current** median would realize
            // if we reassigned that median and all its members to x;
            // note we've already accounted for the fact that the median
            // would save z by closing; now we have to subtract from the savings
            // the extra cost of reassigning that median and its members
            int assign = points->p[i].assign;
            lower[center_table[assign]] += current_cost - x_cost;
        }
    }

    // at this time, we can calculate the cost of opening a center
    // at x; if it is negative, we'll go through with opening it

    for ( int i = 0; i < points->num; i++ ) {
        if ( is_center[i] ) {
            double low = z + work_mem[center_table[i]];
            gl_lower[center_table[i]] = low;
            if ( low > 0 ) {
                // i is a median, and
                // if we were to open x (which we still may not) we'd close i

                // note, we'll ignore the following quantity unless we do open x
                ++number_of_centers_to_close;
                cost_of_opening_x -= low;
            }
        }
    }
    //use the rest of working memory to store the following
    work_mem[K] = number_of_centers_to_close;
    work_mem[K+1] = cost_of_opening_x;

    gl_number_of_centers_to_close = ( int ) work_mem[K];
    gl_cost_of_opening_x = z + work_mem[K+1];

    // Now, check whether opening x would save cost; if so, do it, and
    // otherwise do nothing

    if ( gl_cost_of_opening_x < 0 ) {
        // we'd save money by opening x; we'll do it
        for ( int i = 0; i < points->num; i++ ) {
            bool close_center = gl_lower[center_table[points->p[i].assign]] > 0 ;
            if ( switch_membership[i] || close_center ) {
                // Either i's median (which may be i itself) is closing,
                // or i is closer to x than to its current median
                points->p[i].cost = points->p[i].weight * dist ( points->p[i], points->p[x], points->dim );
                points->p[i].assign = x;
            }
        }
        for ( int i = 0; i < points->num; i++ ) {
            if ( is_center[i] && gl_lower[center_table[i]] > 0 ) {
                is_center[i] = false;
            }
        }
        if ( x >= 0 && x < points->num ) {
            is_center[x] = true;
        }

        *numcenters = *numcenters + 1 - gl_number_of_centers_to_close;
    } khác {
        gl_cost_of_opening_x = 0; // the value we'll return
    }

    free ( work_mem );

    return -gl_cost_of_opening_x;
}

我试图并行化的函数:

/* compute Euclidean distance squared between two points */
float dist ( Point p1, Point p2, int dim )
{
    float result=0.0;
    #pragma omp parallel for reduction(+:result)
    for (int i=0; i
        result += ( p1.coord[i] - p2.coord[i] ) * ( p1.coord[i] - p2.coord[i] );
    }   
    return ( result );
}

重点是:

/* this structure represents a point */
/* these will be passed around to avoid copying coordinates */
Kiểu định nghĩa cấu trúc {
    float weight;
    float *coord;
    long assign; /* number of point where this one is assigned */
    float cost; /* cost of that assignment, weight*distance */
} Point;

我有一个大型的 streamcluster 应用程序(815 行代码)，它生成实时数字并以特定方式对它们进行排序。我在 Linux 上使用过 scalasca 工具，所以我可以测量占用大部分时间的方法，我发现上面列出的方法 dist 是最耗时的。我正在尝试使用 openMP 工具，但并行代码运行的时间比串行代码的运行时间长。如果串行代码在 1.5 秒内运行，则并行化需要 20，但结果是相同的。我想知道是不是由于某种原因我无法并行化这部分代码，或者我没有正确执行。我试图在调用树中对其进行并行化的方法:main->pkmedian->pFL->pgain->dist(-> 表示调用以下方法)

câu trả lời hay nhất

您选择并行化的代码:

float result=0.0;
#pragma omp parallel for reduction(+:result)
for (int i=0; i
    result += ( p1.coord[i] - p2.coord[i] ) * ( p1.coord[i] - p2.coord[i] );
}  

不太适合从并行化中获益。您不应在此处使用 parallel for。您可能不应该在内循环上使用并行化。如果您可以并行化一些外部循环，您会更愿意看到 yield 。

协调线程组启动并行区域会产生开销，之后执行缩减也会产生开销。同时，并行区域的内容基本上不需要运行时间。鉴于此，您需要将 dim 设置得非常大，然后才能期望这会带来性能优势。

为了更形象地表达这一点，请考虑您正在进行的数学运算将花费纳秒并将其与显示各种 OpenMP 指令的开销的图表进行比较。

如果您需要它运行得更快，您的第一站应该是使用适当的编译标志，然后查看 SIMD 操作:SSE 和 AVX 是很好的关键字。您的编译器甚至可能会自动调用它们。

我构建了一些测试代码(见下文)并在启用各种优化的情况下编译它，如下所列，并在 100,000 个元素的数组上运行它。请注意，启用 -O3 会产生与 OpenMP 指令顺序相同的运行时。这意味着在考虑使用 OpenMP 之前，您需要大约 400,000 个数组，为了安全起见，可能更接近 1,000,000 个。

• 没有优化。运行时间约为 1900 微秒。
• -O3:启用许多优化。运行时间约为 200 微秒。
• -ffast-math:你想要这个，除非你正在做一些非常棘手的事情。运行时间大致相同。
• -march=native:编译代码以使用 CPU 的全部功能，而不是可以在许多 CPU 上运行的通用指令集。运行时间约为 100 微秒。

我们开始吧，战略性地使用编译器选项 (-march=native) 可以使相关代码的速度加倍，而无需处理并行问题。

Đây是一个方便的幻灯片演示文稿，其中包含一些解释如何以高性能方式使用 OpenMP 的提示。

Mã kiểm tra:

#include 
#include 
#include 
#include 

int chính(){
  std::vector a;
  std::vector b;
  for(int i=0;i<100000;i++){
    a.push_back(rand()/(double)RAND_MAX);
    b.push_back(rand()/(double)RAND_MAX);
  }

  std::chrono::steady_clock::time_point begin = std::chrono::steady_clock::now();

  float result = 0.0;
  //#pragma omp parallel for reduction(+:result)
  for (unsigned int i=0; i
    result += ( a[i] - b[i] ) * ( a[i] - b[i] );

  std::chrono::steady_clock::time_point end= std::chrono::steady_clock::now();

  std::cout << "Time difference = " << std::chrono::duration_cast(end - begin).count() << " microseconds"<<>
}

关于c++ - OpenMp 并行，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44097890/