Linux IPC实践(12) --System V信号量(2)

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;
}