使用 pragma omp parallel for 计算 pi 的问题

Question

我编写了以下代码来计算 pi 的值并且它有效：

#include <omp.h>
#include <stdio.h>

static long num_steps = 1000000;
double step;
#define NUM_THREADS 16

int main()
{
    int i, nthreads;
    double tdata, pi, sum[NUM_THREADS];
    omp_set_num_threads(NUM_THREADS);
    step = 1.0 / (double)num_steps;

    tdata = omp_get_wtime();

    #pragma omp parallel
    {
        int i, id, nthrds;
        double x;
        id = omp_get_thread_num();
        nthrds = omp_get_num_threads();
        if (id == 0)
            nthreads = nthrds;
        for (i = id, sum[id] = 0.0; i < num_steps; i = i + nthrds)
        {
            x = (i + 0.5) * step;
            sum[id] = sum[id] + 4.0 / (1.0 + x * x);
        }
    }

    tdata = omp_get_wtime() - tdata;
    for (i = 0, pi = 0.0; i < nthreads; i++)
    {
        pi = pi + sum[i] * step;
    }
    printf("pi=%f and it took %f seconds", pi, tdata);
}

然后我了解到我可以使用 #pragma omp parallel for 然后我就不必手动将计算中断到不同的线程。所以我写了这个：

#include <omp.h>
#include <stdio.h>

static long num_steps = 1000000;
double step;
#define NUM_THREADS 16

int main()
{
    int i;
    double tdata, pi, x, sum = 0.0;
    omp_set_num_threads(NUM_THREADS);
    step = 1.0 / (double)num_steps;

    tdata = omp_get_wtime();

    #pragma omp parallel for
    {
        for (i = 0; i < num_steps; i++)
        {
            x = (i + 0.5) * step;
            sum = sum + 4.0 / (1.0 + x * x);
        }
    }

    tdata = omp_get_wtime() - tdata;
    pi = sum * step;
    printf("pi = %f and compute time = %f seconds", pi, tdata);
}

然而，这不起作用并且输出了错误的圆周率值。我究竟做错了什么？谢谢。

Answer 1

您的代码中有两个竞争条件导致了不正确的结果：

评论中已经指出了一个，即变量sum的更新（在线程之间共享 )，可以用reduction子句解决；
另一个是变量x的更新，它也是线程共享的。这种竞争条件可以通过简单地使该变量 private 到线程来解决。

两种可能的解决方案：

使用 OpenMP 的 private 构造函数

 #pragma omp parallel for reduction(+:sum) private(x)
 for (i = 0; i < num_steps; i++)
 {
      x = (i + 0.5) * step;
      sum = sum + 4.0 / (1.0 + x * x);
 }

在范围内声明变量'x'并为

 #pragma omp parallel for reduction(+:sum)
 for (i = 0; i < num_steps; i++)
 {
      double x = (i + 0.5) * step;
      sum = sum + 4.0 / (1.0 + x * x);
 }

最终代码可能如下所示：

    #include <omp.h>
    #include <stdio.h>

    static long num_steps = 1000000;
    #define NUM_THREADS 16

    int main()
    {
        double sum = 0.0;
        omp_set_num_threads(NUM_THREADS);
        double step = 1.0 / (double)num_steps;
    
        double tdata = omp_get_wtime();
    
        #pragma omp parallel for reduction(+:sum)
        for (int i = 0; i < num_steps; i++)
        {
             double x = (i + 0.5) * step;
             sum = sum + 4.0 / (1.0 + x * x);
        }
    
        tdata = omp_get_wtime() - tdata;
        double pi = sum * step;
        printf("pi = %f and compute time = %f seconds", pi, tdata);
    }

使用 pragma omp parallel for 计算 pi 的问题

Problem using pragma omp parallel for to compute pi

c

parallel-processing

multithreading

openmp

thread-safety