Openmp编程练习
火车卖票
生产消费循环队列
#include "stdio.h" #include "omp.h" #include <windows.h> //使用Sleep()函数需要包含此头文件 int buf[5];//缓冲区的大小 int poi; int poi2; int num; omp_lock_t lock; void shengchan() { puts("shengchan"); while (true) { omp_set_lock(&lock); if (num < 5) { while (buf[poi] == 1)poi = (poi + 1) % 5; printf("生产者在%d位置上放置了一个n", poi); buf[poi] = 1; num++; poi = (poi + 1) % 5; } omp_unset_lock(&lock); Sleep(500); } } void xiaofei() { puts("xiaofei"); while (true) { omp_set_lock(&lock); //printf("%dn", num); if (num>=1) { while (buf[poi2] == 0)poi2 = (poi2 + 1) % 5; printf("消费者在%d位置上消费了一个n", poi2); buf[poi2] = 0; num--; } omp_unset_lock(&lock); Sleep(500); } } int main() { omp_init_lock(&lock); #pragma omp parallel sections num_threads(2) { #pragma omp section shengchan(); #pragma omp section xiaofei(); } omp_destroy_lock(&lock); return 0; }
蒙特卡洛圆周率
#include "stdio.h" #include "omp.h" #include <windows.h> //使用Sleep()函数需要包含此头文件 #include<time.h> #include<iostream> using namespace std; double distance(double x, double y) { return sqrt((x - 0.5) * (x - 0.5) + (y - 0.5) * (y - 0.5)); } bool judge(double x,double y) { return distance(x, y) <= 0.5; } int in_num; int main() { /* for (int i = 1; i <= 5; i++) { cout << rand() / (double)RAND_MAX << endl; }*/ bool flag = false; double x; double y; #pragma omp for private(flag,x,y) for (int i = 1; i <= 10000; i++) { x = rand() / (double)RAND_MAX; y = rand() / (double)RAND_MAX; flag = judge(x,y); if (flag) { #pragma omp atomic in_num++; } } double ans = (double)in_num / 10000; cout << ans*4 << endl; }
多线程二维数组和解法1 firstprivate+atomic
#include "stdio.h" #include "omp.h" #include <windows.h> //使用Sleep()函数需要包含此头文件 #include<time.h> #include<iostream> using namespace std; int a[5][5] = { {1,1,1,1,1},{2,2,2,2,2},{3,3,3,3,3},{4,4,4,4,4},{5,5,5,5,5} }; int final_ans = 0; void increase(int temp_sum) { #pragma omp atomic final_ans += temp_sum; } int main() { int temp_sum=0; int i,j; #pragma omp parallel for private(i,j) firstprivate(temp_sum) num_threads(5)//每个线程必须一致,或者采用ppt上的例子进行划分 // firstprivate(temp_sum) reduction(+:temp_sum) 这两个不能同时出现 for (i = 0; i <= 4; i++) { //temp_sum += 1; //printf("%d 当前的temp_sum值为%dn",i, temp_sum); for (j = 0; j <= 4; j++) { temp_sum += a[i][j]; } printf("temp_sum is %dn", temp_sum); increase(temp_sum); } printf("%dn", final_ans); return 0; }
多线程二维数组解法2 线程可以不用对应数量
#include "stdio.h" #include "omp.h" #include <windows.h> //使用Sleep()函数需要包含此头文件 #include<time.h> #include<iostream> using namespace std; int a[5][5] = { {1,1,1,1,1},{2,2,2,2,2},{3,3,3,3,3},{4,4,4,4,4},{5,5,5,5,5} }; int ans_buf[5]; int main() { int i, j; #pragma omp parallel for num_threads(3) private(j) for (int i = 0; i <= 4; i++) { for (int j = 0; j <= 4; j++) { ans_buf[i] += a[i][j]; } } int sum = 0; for (int i = 0; i <= 4; i++) sum += ans_buf[i]; printf("%dn", sum); }
