OpenMP 程序意外地没有因为更多线程而更快

Question

我是 OpenMP 的新手。我编写了一个代码，通过计算函数

f(x) = 4/(1+x^2)

的面积来计算 Pi 的值。我尝试检查不同线程的效率。下面是我的代码：

int main() {
    long long num_iteration = 100000000;
    double pi = 0;
    double x_margin = 1.0 / num_iteration;
    double start_time, end_time;
    int num_threads_main = 0;
    int j = NUM_THREAD;
    int a;

    for(a = 1; a <= j; a++) {
        pi = 0.0;
        omp_set_num_threads(a);
        start_time = omp_get_wtime();

        #pragma omp parallel
        {
            int i;
            int ID = omp_get_thread_num();
            double sum = 0.0;
            int num_threads_local = omp_get_num_threads();
            double x_value = 0;

            if(ID == 0)
                num_threads_main = num_threads_local;

            for (i = ID; i < num_iteration; i = i + num_threads_local) {
                x_value = x_margin * ( i + 0.5 );
                sum = sum + ( 4.0 / (1.0 + x_value * x_value));
            }

            #pragma critical
            pi = pi + sum * x_margin;
        }

        end_time = omp_get_wtime() - start_time;
        printf("pi with 1000000000 steps is %f in %f seconds with threads %d\n", pi, end_time, num_threads_main);
    }

    return 0;
}

然而，当我得到一个奇怪的结果时：

pi with 1000000000 steps is 3.141593 in 0.333257 seconds with threads 1
pi with 1000000000 steps is 3.141593 in 0.175574 seconds with threads 2
pi with 1000000000 steps is 3.141593 in 0.177884 seconds with threads 3
pi with 1000000000 steps is 3.141593 in 0.170591 seconds with threads 4

为什么线程越多我的程序速度反而不快？我知道 OpenMP 可能会产生一些开销。不过，我将我的代码与 Tim Mattson 的 OpenMP 教程代码进行了比较。我发现两者之间非常相似。以下是他的代码：

static long num_steps = 100000000;
double step;
int main ()
{
    int i,j;
    double pi, full_sum = 0.0;
    double start_time, run_time;
    double sum[MAX_THREADS];

    step = 1.0/(double) num_steps;

    for(j=1;j<=MAX_THREADS ;j++){
        omp_set_num_threads(j);
        full_sum = 0.0;
        start_time = omp_get_wtime();

        #pragma omp parallel private(i)
        {
            int id = omp_get_thread_num();
            int numthreads = omp_get_num_threads();
            double x;

            double partial_sum = 0;

            #pragma omp single
            printf(" num_threads = %d",numthreads);

            for (i=id;i< num_steps; i+=numthreads){
                x = (i+0.5)*step;
                partial_sum += + 4.0/(1.0+x*x);
            }

            #pragma omp critical
            full_sum += partial_sum;
        }

        pi = step * full_sum;
        run_time = omp_get_wtime() - start_time;
        printf("\n pi is %f in %f seconds %d threds \n ",pi,run_time,j);
    }
}

这是他的结果：

 num_threads = 1
 pi is 3.141593 in 0.483153 seconds 1 threds 
  num_threads = 2
 pi is 3.141593 in 0.305407 seconds 2 threds 
  num_threads = 3
 pi is 3.141593 in 0.246802 seconds 3 threds 
  num_threads = 4
 pi is 3.141593 in 0.204905 seconds 4 threds 

我在配备 2.3 GHz 双核 Intel Core i5 的 MacOS 10.15.3 上进行了实验。我是不是错过了什么。谢谢你。

Answer 1

您可能没有在 VS 设置中启用 OpenMP。如果不启用它，您仅使用一个线程