Linux IPC Practice (--system v Semaphore) (2)

Practice 1: Signal Volume implementation process mutex

The parent-child process execution process is as follows:

Parent process	Child process
P	P
O (print)	X (print)
Sleep	Sleep
O (print)	X (print)
V	V
Sleep	Sleep

As can be seen, O or X is always paired, either two O, or two x;

/**p,v Primitives implement a parent-child process mutex using terminal **///program code 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)//Sub-process {Srand (getpid ());            while (count--) {sem_p (semid);            Critical section begins cout << ' X '; Fflush (stdout);            Be sure to add ffflush because the interrupt is the row buffer for sleep (rand ()%3);            cout << ' X ';            Fflush (stdout);            Critical section end Sem_v (Semid);        Sleep (rand ()%3);        }} else/parent process {Srand (getpid ());            while (count--) {sem_p (semid);            Critical section begins cout << ' O ';            Fflush (stdout);            Sleep (rand ()%3);            cout << ' O ';            Fflush (stdout);            Critical section end Sem_v (Semid);        Sleep (rand ()%3);        } wait (NULL); Sem_dElete (Semid); } return 0;}

Practice 2: The semaphore set solves the problem of the Philosopher's meal

Suppose there are five philosophers sitting around a round table, doing one of the following two things: eating, or thinking. When they eat, they stop thinking and they stop eating when they think. There is a fork between each of the two philosophers. Since it is difficult to eat with a fork, it is assumed that philosophers must eat with two forks, and that they can only use the two forks on their right and left hand side.

/** solution adopted is: only about two knives and forks can be used, only picked up two knives and forks to achieve a deadlock and no deadlock of the two forms of wait_2fork (see below) **/int semid;//no deadlock 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}}; At the same time get left and right two forks if (Semop (Semid, SOPs, 2) = =-1) err_exit ("Wait_2fork error");}    /*//has a deadlock waitvoid wait_2fork (unsigned short) {unsigned short left = no;    unsigned short right = (no+1)%5;    struct Sembuf SOPs = {left,-1, 0};    Get the left knife and fork if (Semop (Semid, &sops, 1) = =-1) err_exit ("Wait_2fork error");   Sleep (4);    Sleep for a few seconds, accelerating the generation of deadlocks Sops.sem_num = right; Get the right knife and fork if (Semop (Semid, &sops, 1) = =-1) err_exit ("Wait_2fork error");}    *///release two cutlery void signal_2fork (unsigned short) {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");} philosopher 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);        Get two cutlery//dining cout << "+ +" << no << "is eating" << Endl;        Sleep (rand ()%5+1); Signal_2fork (no);//Release two Knives and Forks}}

int main () {    //Creates a semaphore set: Contains 5 semaphores    semid = Semget (ipc_private, 5, ipc_creat|0666);    if (Semid = =-1)        err_exit ("Semget error");    Each semaphore is set to the initial value of 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");    Create four sub-processes, set the number of each process to 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;        }    }    The last five processes (4 sub-processes + 1 parent processes) will be assembled here,    //each process represents a philosopher, ref. No:0~4    philosopher (no);    return 0;}

Linux IPC Practice (--system v Semaphore) (2)