实践1:信号量实现进程互斥
父子进程执行流程如下:
父进程 | 子进程 |
P | P |
O(print) | X(print) |
sleep | sleep |
O(print) | X(print) |
V | V |
sleep | sleep |
从图中可以看出, O或X总是成对出现的, 要么两个O, 要么两个X;
/**P,V原语实现父子进程互斥使用终端**/ // 程序代码 int main(int argc,char *argv[]) { int semid = sem_create(IPC_PRIVATE); sem_setval(semid, 1); int count = 10; pid_t pid = fork(); if (pid == -1) err_exit("fork error"); else if (pid > 0) //子进程 { srand(getpid()); while (count --) { sem_P(semid); //临界区开始 cout << ‘X‘; fflush(stdout); //一定要加上ffflush, 因为中断是行缓冲的 sleep(rand()%3); cout << ‘X‘; fflush(stdout); //临界区结束 sem_V(semid); sleep(rand()%3); } } else //父进程 { srand(getpid()); while (count --) { sem_P(semid); //临界区开始 cout << ‘O‘; fflush(stdout); sleep(rand()%3); cout << ‘O‘; fflush(stdout); //临界区结束 sem_V(semid); sleep(rand()%3); } wait(NULL); sem_delete(semid); } return 0; }
实践2: 信号量集解决哲学家进餐问题
假设有五位哲学家围坐在一张圆形餐桌旁,做以下两件事情之一:吃饭,或者思考。吃东西的时候,他们就停止思考,思考的时候也停止吃东西。每两个哲学家之间有一只餐叉。因为用一只餐叉很难吃饭,所以假设哲学家必须用两只餐叉吃东西, 而且他们只能使用自己左右手边的那两只餐叉。
/** 解决的方法采用的是: 只有左右两个刀叉都能够使用时,才拿起两个刀叉 实现了有死锁和无死锁的两种形式的wait_2fork(见下) **/ int semid; //没有死锁的wait void wait_2fork(unsigned short no) { unsigned short left = no; unsigned short right = (no+1)%5; struct sembuf sops[2] = {{left, -1, 0}, {right, -1, 0}}; //同时获取左右两把刀叉 if (semop(semid, sops, 2) == -1) err_exit("wait_2fork error"); } /* //有死锁的wait void wait_2fork(unsigned short no) { unsigned short left = no; unsigned short right = (no+1)%5; struct sembuf sops = {left, -1, 0}; //获取左边的刀叉 if (semop(semid, &sops, 1) == -1) err_exit("wait_2fork error"); sleep(4); //沉睡几秒, 加速死锁的产生 sops.sem_num = right; //获取右边的刀叉 if (semop(semid, &sops, 1) == -1) err_exit("wait_2fork error"); } */ //释放两把刀叉 void signal_2fork(unsigned short no) { unsigned short left = no; unsigned short right = (no+1)%5; struct sembuf sops[2] = {{left, 1, 0}, {right, 1, 0}}; if (semop(semid, sops, 2) == -1) err_exit("signal_2fork error"); } //哲学家 void philosopher(unsigned short no) { srand(time(NULL)); while (true) { cout << no << " is thinking" << endl; sleep(rand()%5+1); cout << no << " is hunger" << endl; wait_2fork(no); //获取两把刀叉 //进餐 cout << "++ " << no << " is eating" << endl; sleep(rand()%5+1); signal_2fork(no);//释放两把刀叉 } }
int main() { // 创建一个信号量集: 里面包含5个信号量 semid = semget(IPC_PRIVATE, 5, IPC_CREAT|0666); if (semid == -1) err_exit("semget error"); //将每个信号量都设初值为1 union semun su; su.val = 1; for (int i = 0; i < 5; ++i) if (semctl(semid, i, SETVAL, su) == -1) err_exit("semctl SETVAL error"); //创建四个子进程, 将每个进程的编号设定为no pid_t pid; unsigned short no = 0; for (unsigned short i = 0; i < 4; ++i) { pid = fork(); if (pid == -1) err_exit("fork error"); else if (pid == 0) { no = i+1; break; } } // 最后五个进程(4个子进程+1个父进程)都会汇集到此处, // 每个进程代表着一个哲学家,编号no: 0~4 philosopher(no); return 0; }
Linux IPC实践(12) --System V信号量(2)
原文:http://blog.csdn.net/zjf280441589/article/details/43882881