共享名为POSIX的信号量 [英] Share named POSIX semaphores

问题描述

我在理解如何在多个进程之间共享POSIX信号时遇到问题.我正在尝试执行以下操作:
1.生产者初始化一个信号量
2.生产者将10个令牌发布到该信号量，并在此之前睡眠1秒
3.消费者从信号量中获取令牌
当我启动生产者时，出现分段错误(核心已转储).此外，我不确定共享命名信号量的方式是否正确.
生产者:

I have issues understanding how to share a POSIX semaphore among multiple processes. I am trying to do the following:
1. The producer initializes a semaphore
2. The producer posts 10 tokens to the semaphore and sleeps 1 second before doing so
3. The consumer gets a token from the semaphore
When I start my producer, a segmentation fault (core dumped) occurs. Furthermore I am not sure, if my way of sharing the named semaphore is correct.
Producer:

#include <semaphore.h>
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/mman.h>

#define SEM_NAME "/mutex"

int main () {
    sem_t* sem = sem_open(SEM_NAME,O_CREAT,0644,0);
    for (int i = 0; i<10; i++) {
        sleep(1);
        sem_post(sem);
        printf("Token was posted! \n");
    }   
    sem_close(sem);
    sem_unlink(SEM_NAME);
}

消费者:

#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#include <semaphore.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/syscall.h>
#include <fcntl.h>    

int main () {
    sem_t *mutex = sem_open("/mutex",O_CREAT);
    for(int i = 0; i<10; i++) {
        sem_wait(mutex);
        printf("One Token was consumed! %d",(int) getpid());
    }
    sem_close(mutex);
}

推荐答案

让您的消费者等待:

sem_wait(mutex);

并冲洗每张照片(如果没有打印，可能会在末尾全部冲洗掉):

and flush each print (if not prints may all be flushed at the end):

print("One token consumed\n");

也；请注意从打开处返回的值:

Also; please take care of returned value from open:

if (mutex==SEM_FAILED) exit(1);

和

if (sem==SEM_FAILED) exit(1);

这篇关于共享名为POSIX的信号量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！