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我目前正在学习 POSIX 线程（pthread）。

我现在创建了一个简单的程序，它将共享值增加 7 直到超过 10000，然后它应该向下一个线程发出一个条件信号，将其减少 3 直到低于 1000。最后它应该将结果除以 2，并且 main 应该输出结果。

我的代码：

pthread_t threads[3];
pthread_cond_t cond_a, cond_b;
pthread_mutex_t mutex;

int counter;

void * worker_one();
void * worker_two();
void * worker_three();

int main(int argv, const char ** argc) {
    counter = 0;

    pthread_cond_init(&cond_a, NULL);
    pthread_cond_init(&cond_b, NULL);
    pthread_mutex_init(&mutex, NULL);

    pthread_create(&threads[0], NULL, worker_one, NULL);
    pthread_create(&threads[1], NULL, worker_two, NULL);
    pthread_create(&threads[2], NULL, worker_three, NULL);

    pthread_join(threads[0], NULL);
    pthread_join(threads[1], NULL);
    pthread_join(threads[2], NULL);

    printf("Value started at %d and ends with %d.\n", 0, counter);

    return 0;
}

void * worker_one() {
    printf("Worker one started.\n");

    pthread_mutex_lock(&mutex);

    printf("Worker one starting work.\n");
    while (counter < 10000) {
        counter += 7;
    }

    pthread_cond_signal(&cond_a);

    printf("Worker one finished work with: %d.\n", counter);

    pthread_mutex_unlock(&mutex);
    pthread_exit(NULL);
}

void * worker_two() {
    printf("Worker two started.\n");

    pthread_mutex_lock(&mutex);
    pthread_cond_wait(&cond_a, &mutex);

    printf("Worker two starting work.\n");
    while (counter > 1000)
        counter -= 3;

    printf("Worker two finished work with: %d.\n", counter);

    pthread_cond_signal(&cond_b);
    pthread_mutex_unlock(&mutex);

    sleep(1);

    pthread_exit(NULL);
}

void * worker_three() {
    printf("Worker three started.\n");

    pthread_mutex_lock(&mutex);
    pthread_cond_wait(&cond_b, &mutex);

    printf("Worker three starting work.\n");

    counter /= 2;

    printf("Worker three finished work with: %d.\n", counter);

    pthread_mutex_unlock(&mutex);
    pthread_exit(NULL);
}

由于某种原因，整个执行都围绕第一个线程进行。信号也被触发，但线程二没有反应。

有人可以告诉我我做错了什么吗？

4

1 回答 1

6

我在这里回答了一个类似的问题：Linux 上的 pthread 条件变量，奇怪的行为。

问题是您甚至在测试您想要等待的条件为真之前就等待。发生的情况是线程 1 在线程 2 等待之前发出信号，因此信号丢失，线程 2 将永远等待。

为了避免这种情况，首先测试您要等待的内容，然后仅在它不在这里时才等待。

编辑：好的，这是一个可能的解决方案，只有一个互斥锁和一个条件（未经测试）

线程 1：

pthread_mutex_lock(&mutex); 
while(thread_1_should_work == false){ // wait until the condition is satisfied
  pthread_cond_wait(&cond, &mutex); 
}

//at this point, we owe the mutex and we know thread_1_should_work is true; 

// do work 

thread_1_shoudl_work = false; 
thread_2_should_work = true; 

pthread_cond_broadcast(&cond); //wake up any waiting thread (if it's not their turn, they'll call wait again)
pthread_mutex_unlock(&mutex);

... 等等。

于 2013-05-29T16:31:21.677 回答