pthead cond的理解

pthread_cond_wait/pthread_cond_timedwait返回时候需要得到mutex，因此有如下两个问题
－对于pthread_cond_timedwait()的理解：到了timeout，下面的程序一定能执行吗？
  答案应该是否定的，pthread_cond_timedwait()也需要获得mutex，否则也不能唤醒
－pthread_cond_signal()只能唤醒一个，如果有多个等待，则不能唤醒
  pthread_cond_broadcast()能唤醒多个，但是通常需要等待mutex按照顺序一个一个进行下面处理

因此在多线程的程序设计中要注意任何一个线程都不能长时间占用资源

下面有几个例子

例子1，时间到，但是timedwait不能即使返回

#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <errno.h>
#include <sys/time.h>

void *fun1(void *arg);
void *fun3(void *arg);

pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
int a = 10;
int b = 5;

int main(void)
{
    printf("hello\n");

    pthread_t id1;
    pthread_create(&id1, NULL, fun1, NULL);
    pthread_t id3;
    pthread_create(&id3, NULL, fun3, NULL);

    pthread_join(id1, NULL);
    pthread_join(id3, NULL);
    printf("main exit\n");

    return 0;
}

void *fun1(void *arg)
{   
    printf("now in the fun1\n");
    pthread_mutex_lock(&mutex);
    printf("fun1 lock\n");
    struct timeval now;
    struct timespec timeout;
    int retcode = 0;

    while (a > b) {
        gettimeofday(&now, NULL);
        timeout.tv_sec = now.tv_sec + 1;
        timeout.tv_nsec = now.tv_usec * 1000;
        retcode = pthread_cond_timedwait(&cond, &mutex, &timeout);
        printf("---fun1 timedwait end\n");
    }
    printf("fun1 will unlock\n");
    pthread_mutex_unlock(&mutex);
    
    return NULL;
}   
void *fun3(void *arg)
{
    printf("now in the fun3\n");
    sleep(3);
    pthread_mutex_lock(&mutex);
    printf("fun3 lock\n");
    a = 3;
    pthread_cond_broadcast(&cond);
    printf("fun3 broadcast and wait 10s\n");
    sleep(10);
    printf("fun3 will unlock\n");
    pthread_mutex_unlock(&mutex);
    return NULL;
}

执行结果

hello
now in the fun1
fun1 lock
now in the fun3
---fun1 timedwait end
---fun1 timedwait end
fun3 lock
fun3 broadcast and wait 10s
fun3 will unlock
---fun1 timedwait end
fun1 will unlock
main exit

例子2 cond_signal不能唤醒多个 （修改下面的注释，可以展示另外一个问题）

#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <errno.h>

void *fun1(void *arg);
void *fun2(void *arg);
void *fun3(void *arg);

pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
int a = 10;
int b = 5;

int main(void)
{
    printf("hello\n");

    pthread_t id1;
    pthread_create(&id1, NULL, fun1, NULL);
    pthread_t id2;
    pthread_create(&id2, NULL, fun2, NULL);
    pthread_t id3;
    pthread_create(&id3, NULL, fun3, NULL);

    pthread_join(id1, NULL);
    pthread_join(id2, NULL);
    pthread_join(id3, NULL);
    printf("main exit\n");

    return 0;
}

void *fun1(void *arg)
{
    printf("now in the fun1\n");
    pthread_mutex_lock(&mutex);
    printf("fun1 lock\n");
    while (a > b)
        pthread_cond_wait(&cond, &mutex);
    //printf("fun1 end wait\n");
    //sleep(10);
    printf("fun1 will unlock\n");
    pthread_mutex_unlock(&mutex);

    return NULL;
}
void *fun2(void *arg)
{
    printf("now in the fun2\n");
    pthread_mutex_lock(&mutex);
    printf("fun2 lock\n");
    while (a > b)
        pthread_cond_wait(&cond, &mutex);
    //printf("fun2 end wait\n");
    //sleep(10);
    printf("fun2 will unlock\n");
    pthread_mutex_unlock(&mutex);

    return NULL;
}
void *fun3(void *arg)
{
    printf("now in the fun3\n");
    sleep(3);
    pthread_mutex_lock(&mutex);
    printf("fun3 lock\n");
    a = 3;
    //pthread_cond_broadcast(&cond);
    pthread_cond_signal(&cond);
    printf("fun3 will unlock\n");
    pthread_mutex_unlock(&mutex);
    return NULL;
}

结果：其中一个线程不能退出（这里是fun2)

hello
now in the fun1
now in the fun3
now in the fun2
fun1 lock
fun2 lock
fun3 lock
fun3 will unlock
fun1 will unlock