sem_wait没有基本的code工作

问题描述

用gcc编译。我跑这来看看为什么我用我的其他程序都信号灯无法正常工作。我只是用这些不当或什么？该字符串输出每次即使信号灯应停止执行，并导致死锁，对不对？

Compiled with gcc. I ran this to see why the semaphores I was using in my other programs were not working correctly. Am I just using these them incorrectly or what? The string is outputted every time even though the semaphore should halt execution and cause a deadlock, right?

下面是code：

#include <pthread.h>
#include <semaphore.h>
#include <stdlib.h>
#define NUM_THREADS 5
void printHello();
int main(){
    int i;
    pthread_t threads[NUM_THREADS];
    sem_t sem1;
    sem_init(&sem1, 0, 0);
    sem_wait(&sem1);

    for(i = 0; i < NUM_THREADS; i++){
        pthread_create(&threads[i], NULL, &printHello, NULL);
    }

    sem_destroy(&sem1);
    pthread_exit(NULL);

        return 0;
}

void printHello(){
    printf("sem_wait failed\n");
}

任何帮助将大大AP preciated因为我试图把握整个概念的多线程

Any help would be greatly appreciated as I am trying to grasp the whole multithreading concept.

谢谢！

推荐答案

这code看起来不错。那么，它会更改启动功能后无效* printHello（无效*）和在pthread_join 让你不ŧ破坏信号灯和退出。但信号块，正是因为它应该在Linux上。

That code looks fine. Well, it will after you change the start function to void *printHello(void *) and pthread_join so you don't destroy the semaphore and exit. But the semaphore blocks exactly as it should on linux.

您在运行OSX？显然OSX不支持POSIX无名信号量。如果是这样的话，你需要使用一个名为POSIX信号即 sem_open ，而不是 sem_init 。

Are you running on OSX? Apparently OSX does not support unnamed POSIX semaphores. If that is the case you need to use named POSIX semaphores i.e. sem_open rather than sem_init.

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