Linux IPC实践(12) --System V信号量(2)

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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

O(print)

X(print)

sleep

O(print)

X(print)

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waitvoid 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waitvoid 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;}

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