【openMP並行計算】計算π

本系列文章均爲個人學習筆記

Linux環境對OpenMP的支持：
在Linux上編譯和運行OpenMP程序
編譯OpenMP程序： gcc -fopenmp a.c
運行OpenMP程序： ./a.out

1. 串行計算π

#include <stdio.h>
#include <omp.h>
static long num_steps = 100000000;
double step;
int main ()
{
	  int i;
	  double x, pi, sum = 0.0;
	  double start_time, run_time;

	  step = 1.0/(double) num_steps;

        	 
	  start_time = omp_get_wtime();

	  for (i=1;i<= num_steps; i++){
		  x = (i-0.5)*step;
		  sum = sum + 4.0/(1.0+x*x);
	  }

	  pi = step * sum;
	  run_time = omp_get_wtime() - start_time;
	  printf("\n pi with %ld steps is %lf in %lf seconds\n ",num_steps,pi,run_time);
}	  

結果：

2. 使用並行域並行化的程序:

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

#define MAX_THREADS 4

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 //並行域開始，每個線程(0和1)都會執行該代碼
      {
         int i;
         int id = omp_get_thread_num();
         int numthreads = omp_get_num_threads();
         double x;

         sum[id] = 0.0;

         if (id == 0) //保證只有一個線程輸出ID
            printf(" num_threads = %d",numthreads);

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

      for(full_sum = 0.0, i=0;i<j;i++){
         full_sum += sum[i];
      }

      pi = step * full_sum;
      run_time = omp_get_wtime() - start_time;
      printf("\n pi is %f in %f seconds %d thrds \n",pi,run_time,j);
   }
}	
//共4個線程參加計算，其中線程0進行迭代步0,4,...線程1進行迭代步1,5,....  

結果：

我們發現使用並行計算結果耗時更多，因爲 False sharing。（自行百度）

3.private字句和critical制導語句並行化：

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

#define MAX_THREADS 4

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) //該子句表示 i 變量對於每個線程是私有的
		{
			int id = omp_get_thread_num();
			int numthreads = omp_get_num_threads();
			double x;

			double partial_sum = 0;

			#pragma omp single	//作用和上一份代碼一樣，保證只有一個線程輸出ID
				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	//指定代碼段在同一時刻只能由一個線程進行執行
				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);
	}
}
//共4個線程參加計算，其中線程0進行迭代步0,4....，線程1進行迭代步1,5....

結果;

速度明顯快了很多！

4.並行規約並行化:

#include <stdio.h>
#include <omp.h>
static long num_steps = 100000000;
double step;
int main ()
{
	int i;
	double x, pi, sum = 0.0;
	double start_time, run_time;

	step = 1.0/(double) num_steps;
	
	for (i = 1; i <= 4; i++)
	{
		sum = 0.0;
		omp_set_num_threads(i);
		start_time = omp_get_wtime();

#pragma omp parallel  
{
#pragma omp single	
		printf(" num_threads = %d",omp_get_num_threads());

#pragma omp for reduction(+:sum)	//每個線程保留一份私有拷貝sum,最後對線程中所以sum進行+規約，並更新sum的全局值
		for (i=1;i<= num_steps; i++){
			x = (i-0.5)*step;
			sum = sum + 4.0/(1.0+x*x);
		}
}
		pi = step * sum;
		run_time = omp_get_wtime() - start_time;
		printf("\n pi is %f in %f seconds and %d threads\n",pi,run_time,i);
	}
}	
//共4個線程參加計算，其中線程0進行迭代步0~24999，線程1進行迭代步24999~50000.  

結果：

速度一般般