【问题标题】:Building semaphores using Mutexes使用互斥体构建信号量
【发布时间】:2021-11-18 01:12:18
【问题描述】:

我正在阅读现代操作系统,有一个练习要求您使用二进制信号量构建计数信号量。

我想知道是否可以使用信号量和共享内存(C 和 POSIX IPC)来做到这一点。

我将在此处简化我的程序。

typedef struct sem {
    int semid; /* contains two semaphores with value set to 1, one for the critical region and the other for locking */
    int shmid; /* shared counter */
}Sem;

/* the parent process spawns a few child processes and then waits for them */
void child() {
    Sem *s;
    
    s = get_sem();
    down_sem(s);
    printf("working...");
    sleep(5);
    printf("finished!");
    up_sem(s);
}

void down_sem(Sem *s) {
    int n;

    enter_critical(s); // down to critical region sem
    n = get_counter(s);
    
    if (n == 0) {
        exit_critical(s); // up to critical region sem
        acquire_lock(s); // down to lock sem
    } else {
        n--;
        exit_critical(s); // up to critical region sem
    }
}

void up_sem(Sem *s) {
    int n;

    enter_critical(s);
    n = get_counter(s);
    n++;
    
    if (n == 1) {
        exit_critical(s);
        release_lock(s); // up to lock sem
    } else {
        exit_critical(s);
    }
}

但这不起作用,因为某些进程永远阻塞在锁信号量中。我现在才刚刚学习这些概念,所以我的设计可能完全偏离了。

如果你愿意,我可以分享完整的代码。

提前致谢!

编辑:请求mre

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>
#include <sys/wait.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <sys/types.h>
#include <sys/shm.h>

#define NPROC 5
#define SLOTS 2

#define PATH "."
#define SID 'S'
#define SEMN 2

#define SHMID 'M'
#define SHM_SIZE 32

#define CRIT 0
#define LOCK 1

#define UP 1
#define DOWN -1

typedef struct sem {
    int shmid;
    int semid;
}Sem;

typedef union semun {
    int val;
    struct semid_ds *buf;
    unsigned short *array;
}Args;

void err (char *msg) {
    char error[50];
    int n;

    n = snprintf(error, 50, "[%d] %s", getpid(), msg);
    error[n] = 0;
    perror(error);
    exit(1);
}

int* get_n(Sem *s) {
    int *data = NULL;

    data = shmat(s->shmid, (void *)0, 0);
    if (data == (int *)(-1))
        err("get_n shmat");

    return data;
}

void dettach_n(int *data) {
    if (shmdt(data) == -1)  
        err("dettach_n shmdt");
}

void enter_crit(Sem *s) {       
    struct sembuf sb;

    sb.sem_num = CRIT;
    sb.sem_op = DOWN;

    printf("[%d] enter crit\n", getpid());
    if (semop(s->semid, &sb, 1) == -1)
        err("enter crit");
}

void exit_crit(Sem *s) {
    struct sembuf sb;

    sb.sem_num = CRIT;
    sb.sem_op = UP;

    printf("[%d] exit crit\n", getpid());
    if (semop(s->semid, &sb, 1) == -1)
        err("exit crit");
}

void acquire_lock(Sem *s) {
    struct sembuf sb;

    sb.sem_num = LOCK;
    sb.sem_op = DOWN;

    printf("[%d] acquire lock\n", getpid());
    if (semop(s->semid, &sb, 1) == -1)
        err("acquire lock");
}

void release_lock(Sem *s) {
    struct sembuf sb;

    sb.sem_num = LOCK;
    sb.sem_op = UP;

    printf("[%d] release lock\n", getpid());
    if (semop(s->semid, &sb, 1) == -1)
        err("release lock");
}

int sem_init(Sem *s, int n) {
    key_t key;
    Args arg;
    int *data;

    /* sem */
    if ((key = ftok(PATH, SID)) == -1) {
        return -1;
    }
    if ((s->semid = semget(key, SEMN, 0600 | IPC_CREAT)) == -1) {
        return -1;
    }

    arg.val = 1;
    if (semctl(s->semid, 0, SETVAL, arg) == -1) {
        return -1;
    }
    if (semctl(s->semid, 1, SETVAL, arg) == -1) {
        return -1;
    }

    /* mem */
    if ((key = ftok(PATH, SHMID)) == -1) {
        return -1;
    }
    if ((s->shmid = shmget(key, SHM_SIZE, 0664 | IPC_CREAT)) == -1) {
        return -1;
    }

    data = shmat(s->shmid, (void *)0, 0);
    if (data == (int *)(-1)) {
        return -1;
    }
    *data = n;
    if (shmdt(data) == -1) {
        return -1;
    }

    return 0;
}

int sem_get(Sem *s) {
    key_t key;
    
    if ((key = ftok(PATH, SID)) == -1) {
        return -1;
    }
    if ((s->semid = semget(key, SEMN, 0)) == -1) {
        return -1;
    }

    if ((key = ftok(PATH, SHMID)) == -1) {
        return -1;
    }
    if ((s->shmid = shmget(key, SHM_SIZE, 0)) == -1) {
        return -1;
    }

    return 0;
}

int sem_up(Sem *s) {
    int *data;

    enter_crit(s);
    data = get_n(s);
    printf("[%d] read %d\n", getpid(), *data);
    (*data)++;

    if (*data == 1) {
        printf("[%d] now is %d\n", getpid(), *data);
        dettach_n(data);
        exit_crit(s);
        release_lock(s);
    } else {
        exit_crit(s);
    }

    return 0;
}

int sem_down(Sem *s) {
    int *data;

    enter_crit(s);
    data = get_n(s);

    printf("[%d] checked %d\n", getpid(), *data);
    if (*data == 0) {
        dettach_n(data);
        exit_crit(s);
        acquire_lock(s);
    } else {
        (*data)--;
        dettach_n(data);
        exit_crit(s);
    }

    return 0;
}

int sem_rm(Sem *s) {
    if (semctl(s->semid, 0, IPC_RMID) == -1)
        return -1;

    if (shmctl(s->shmid, 0, IPC_RMID) == -1)
        return -1;

    return 0;
}

void child() {
    pid_t pid;
    Sem s;

    pid = getpid();
    printf("\x1b[31m[%d] hello!\033[0m\n", pid);

    if (sem_get(&s) == -1) {
        perror("sem_get");
        exit(1);
    }   
    sem_down(&s);
    printf("\x1b[32m[%d] working...\033[0m\n", pid);
    sleep(5);
    printf("\x1b[32m[%d] finishing...\033[0m\n", pid);
    sem_up(&s);
    printf("\x1b[34m[%d] **child leaving**\033[0m\n", pid);
    exit(0);
}

int main() {
    Sem s;
    int i;

    if (sem_init(&s, SLOTS) == -1) {
        perror("sem_init");
        exit(1);
    }

    printf("[%d] parent\n", getpid());

    for (i = 0; i < NPROC; i++) {
        switch(fork()) {
        case 0:
            child();
        case -1:
            perror("fork");
            exit(1);
        }
    }   

    printf("waiting for children...\n");
    for (i = 0; i < NPROC; i++)
        wait(NULL);

    if (sem_rm(&s) == -1) {
        perror("sem_rm");
        exit(1);
    }

    printf("good bye!\n");

    return 0;
}

【问题讨论】:

  • 您可能需要标记此C 以获得更广泛的受众。
  • 不知道。我从未使用过“二进制信号量”,也从未见过任何需要。当一个线程在它已经有一个单元的情况下试图发出这样一个信号量时会发生什么?是信号线程阻塞还是信号单元“丢失”?
  • 您对该程序的简化近似没有特别帮助。我们通常的请求是minimal reproducible example,我们似乎很可能需要一个。
  • @JohnBollinger 那里

标签: c linux operating-system mutex semaphore


【解决方案1】:

问题出在 up_sem 函数中。如果 n 之前为 0,它只执行锁定 sem (release_lock func)。这是不正确的,因为它可能有多个进程在等待。

我的解决方案是添加一个额外的共享内存来计算等待进程,并在该数字大于 0 时执行锁定 sem。

【讨论】:

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