OpenMP

• 包含头文件

  #include<omp.h>

• 声明并行段开始

  #pragma omp paralled;
  {
    int ID=omp_get_thread_num();
    printf("hello(%d)\n",ID);
    printf("world(%d)\n",ID)
  }

• 编译运行

  gcc paralle.c -fopenmp -o paralle

• 完整代码结构

    #include <stdio.h>
    #include <stdlib.h>
    #include <omp.h>
    int main(){
      int ID,nthread;
      omp_set_num_threads(2);
      #pragma omp parallel 
      {
        ID=omp_get_thread_num(); //获取每个线程的id编号
        printf("hello(%d)\n",ID);
        printf("world(%d)\n",ID);
        if(ID==0){
          nthread=omp_get_num_threads(); //获取开辟的线程数目，主进程的线程id默认为0
          printf("Number of threads %d\n",nthread);
        }
      }
      return 0;
    }

• 输出结果

  hello(1)
  world(1)
  hello(0)
  world(1)

:waring: 这是由于线程编号ID定义在公共内存区域，当进行并发时两个线程同时进行默认是ID=0的主线程先执行；所以当执行到if判断语句时，两个线程对ID的赋值语句都已经结束；所以不会进入if语句

• 改变 ID的作用域

  当将ID放在parallel作用域时，只对单个线程可见；所用最终结果会进入到if判断语句

    hello(0)
    world(0)
    Number of threads 2
    hello(1)
    world(1)

• 使用并发求PI

  #include <time.h>
  #include <stdio.h>
  #include <stdlib.h>
  #include "omp.h"
  #define NUM_THREADS 5
  //#define NUM_THREADS atoi(getenv("THREAD")) 使用环境变量进行声明
  #define PAD 8 //跟CPU内存分配有关
  double step;
  static long num_steps;
  int main()
  {
    clock_t start_t, end_t;
    double total_t;
    start_t = clock();
    num_steps = 100000000;
    double  pi, sum[NUM_THREADS][PAD];
    pi=0.0;
    step = 1.0 / (double)num_steps; //将1平分成100000步
    omp_set_num_threads(NUM_THREADS); //设置要使用的线程数目
  #pragma omp parallel  //开始并发执行
    {
      double x;
      int id , i;
      id = omp_get_thread_num(); //获取每个线程的id编号
      sum[id][0] = 0.0;
      for (i=id; i < num_steps; i = i + NUM_THREADS)
      {
        x = (i - 0.5) * step; 
        sum[id][0] += 4.0 / (1.0 + x * x);
      }
    }
      int i;
    for ( i = 0; i < NUM_THREADS; i++)
    {
      pi += sum[i][0] * step;
    }
    end_t = clock();
    total_t = (double)(end_t - start_t) / CLOCKS_PER_SEC;
    printf("运行时间为%fs\t%f\n", total_t, pi);
    return 0;
  }

PAD模式代码

• NUM_THREADS为对应的线程数

• num_steps 为步长

• 数组的读取可以同步进行,没一行是一个数组指针

#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include "omp.h"
#define NUM_THREADS 6
#define PAD 8 //跟CPU内存分配有关
double step;
static long num_steps;
int main()
{
  clock_t start_t, end_t;
  double total_t;
  start_t = clock();
  num_steps = 100000000;
  double  pi, sum[NUM_THREADS][PAD];
  pi=0.0;
  step = 1.0 / (double)num_steps; //将1平分成100000步
  omp_set_num_threads(NUM_THREADS); //设置要使用的线程数目
#pragma omp parallel  //开始并发执行
  {
    double x;
    int id , i;
    id = omp_get_thread_num(); //获取每个线程的id编号
    sum[id][0] = 0.0;
    for (i=id; i < num_steps; i = i + NUM_THREADS)
    {
      x = (i - 0.5) * step; 
      sum[id][0] += 4.0 / (1.0 + x * x);
    }
  }
int i;
  for (i = 0; i < NUM_THREADS; i++)
  {
    pi += sum[i][0] * step;
  }
  end_t = clock();
  total_t = (double)(end_t - start_t) / CLOCKS_PER_SEC;
  printf("运行时间为%fs\t%f\n", total_t, pi);
  return 0;
}

并行域代码

• 伪共享模式，当使用一纬数组时，数组的读取被锁定，因为数组只有一个数组指针

#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include "omp.h"
#define NUM_THREADS 4
#define PAD 8 //跟CPU内存分配有关
double step;
static long num_steps;
int main()
{
  clock_t start_t, end_t;
  double total_t;
  start_t = clock();
  num_steps = 100000000;
  double  pi, sum[NUM_THREADS];
  pi=0.0;
  step = 1.0 / (double)num_steps; //将1平分成100000步
  omp_set_num_threads(NUM_THREADS); //设置要使用的线程数目
#pragma omp parallel  //开始并发执行
  {
    double x;
    int id , i;
    id = omp_get_thread_num(); //获取每个线程的id编号
    sum[id]= 0.0;
    for (i=id; i < num_steps; i = i + NUM_THREADS)
    {
      x = (i - 0.5) * step; 
      sum[id] += 4.0 / (1.0 + x * x);
    }
  }
int i;
  for (i = 0; i < NUM_THREADS; i++)
  {
    pi += sum[i] * step;
  }
  end_t = clock();
  total_t = (double)(end_t - start_t) / CLOCKS_PER_SEC;
  printf("运行时间为%fs\t%f\n", total_t, pi);
  return 0;
}

窜行模式

• 输入同样的步长进行比较

#include <time.h>
#include <stdio.h>
double step;
int main()
{
    static long num_steps;
    // printf("请输入一个1000以上的数字求取PI值\n>>");
    //scanf("%ld",&num_steps);
    for (num_steps = 100000000; num_steps <= 100000000; num_steps += 100000)
    {
        int index;
        index = ((double)num_steps - 1000000) / 100000 + 4;
        clock_t start_t, end_t;
        double total_t;
        int i;
        double x, pi, sum = 0.0;
        start_t = clock();
        step = 1.0 / (double)num_steps; //将1平分成100000步
        for (i = 1; i <= num_steps; i++)
        {
            x = (i - 0.5) * step; //获得0-1的连续值
            sum += 4.0 / (1.0 + x * x);
        }
        pi = step * sum; //
        end_t = clock();
        total_t = (double)(end_t - start_t) / CLOCKS_PER_SEC;
        printf("运行时间为%fs\t%.*f\r\n", total_t, pi, index);
    }
    return 0;
}