多线程互斥与同步

[PonyCheng2019]

简单的写个模拟生产者消费者的多线程代码，
生产者：每次对一个全局变量加1，并唤醒消费者输出该变量的值
消费者：每次等待生产者对全局变量加1，之后输出全局变量的值
期望是消费者每次从1开始，依次打印全局变量的值。
代码如下：

#include <stdio.h>

#include <unistd.h>
#include <pthread.h>

int val;
pthread_mutex_t val_lock;
pthread_cond_t  val_cond;

void *producer_thread(void *args)
{
        while (1) {
                pthread_mutex_lock(&val_lock);
                val++;
                pthread_cond_signal(&val_cond);
                pthread_mutex_unlock(&val_lock);
                sleep(1);
        }
        return NULL;
}

void *consumer_thread(void *args)
{
        while (1) {
                pthread_mutex_lock(&val_lock);
                pthread_cond_wait(&val_cond, &val_lock);
                printf("consumer_thread woke up, val is: %d\n", val);
                pthread_mutex_unlock(&val_lock);
                sleep(1);
        }
        return NULL;
}

int main(int argc, const char *argv[])
{
        pthread_t producer_tid;
        pthread_t consumer_tid;

        pthread_mutex_init(&val_lock, NULL);
        pthread_cond_init(&val_cond, NULL);

        pthread_create(&producer_tid, NULL, producer_thread, NULL);
        pthread_create(&consumer_tid, NULL, consumer_thread, NULL);

        pthread_mutex_destroy(&val_lock);
        pthread_cond_destroy(&val_cond);

        pthread_join(producer_tid, NULL);
        pthread_join(consumer_tid, NULL);

        while (1)
                sleep(1);

        return 0;
}

可是实际的效果如下：
consumer_thread woke up, val is: 1
consumer_thread woke up, val is: 2
consumer_thread woke up, val is: 2
consumer_thread woke up, val is: 4
consumer_thread woke up, val is: 5
consumer_thread woke up, val is: 6

疑问：
1、这段代码处理线程互斥和同步的用法对吗？特别是对条件变量的用法。
2、为什么红色区域的输出会有两次的值都是2呢？

[金秋枣香]

用法不对，生产者应该这样 pthread_mutex_lock(&val_lock); val++; pthread_mutex_unlock(&val_lock); pthread_cond_signal(&val_cond); 先解锁，再去发送信号....

[zhxianbin]

打印之前判断是否还是上次的值不就行了