LINUX环境下多线程编程确定会遇到须要条件变量的状况,此时必然要使用pthread_cond_wait()函数。但这个函数的执行过程比较难于理解。
pthread_cond_wait()的工做流程以下(以MAN中的EXAMPLE为例):
Consider two shared variables x and y, protected by the mutex mut, and a condition vari-
able cond that is to be signaled whenever x becomes greater than y.html
int x,y;
pthread_mutex_t mut = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;linux
Waiting until x is greater than y is performed as follows:编程
pthread_mutex_lock(&mut);
while (x <= y) {
pthread_cond_wait(&cond, &mut);
}
/* operate on x and y */
pthread_mutex_unlock(&mut);多线程
Modifications on x and y that may cause x to become greater than y should signal the con-
dition if needed:app
pthread_mutex_lock(&mut);
/* modify x and y */
if (x > y) pthread_cond_broadcast(&cond);
pthread_mutex_unlock(&mut);ide
这个例子的意思是,两个线程要修改X和 Y的值,第一个线程当X<=Y时就挂起,直到X>Y时才继续执行(由第二个线程可能会修改X,Y的值,当X>Y时唤醒第一个线程),即 首先初始化一个普通互斥量mut和一个条件变量cond。以后分别在两个线程中分别执行以下函数体:函数
pthread_mutex_lock(&mut);
while (x <= y) {
pthread_cond_wait(&cond, &mut);
}
/* operate on x and y */
pthread_mutex_unlock(&mut);oop
和: pthread_mutex_lock(&mut);
/* modify x and y */
if (x > y) pthread_cond_signal(&cond);
pthread_mutex_unlock(&mut);
其实函数的执行过程很是简单,在第一个线程执行到pthread_cond_wait(&cond,&mut)时,此时若是X<=Y,则此函数就将mut互斥量解锁 ,再将cond条件变量加锁 ,此时第一个线程挂起 (不占用任何CPU周期)。
而在第二个线程中,原本由于mut被第一个线程锁住而阻塞,此时由于mut已经释放,因此能够得到锁mut,而且进行修改X和Y的值,在修改以后,一个IF语句断定是否是X>Y,若是是,则此时pthread_cond_signal()函数会唤醒第一个线程 ,并在下一句中释放互斥量mut。而后第一个线程开始从pthread_cond_wait()执行,首先要再次锁mut , 若是锁成功,再进行条件的判断 (至于为何用WHILE,即在被唤醒以后还要再判断,后面有缘由分析),若是知足条件,则被唤醒 进行处理,最后释放互斥量mut 。性能
至于为何在被唤醒以后还要再次进行条件判断(即为何要使用while循环来判断条件),是由于可能有“惊群效应”。有人以为此处既然是被唤醒的,确定 是知足条件了,其实否则。若是是多个线程都在等待这个条件,而同时只能有一个线程进行处理,此时就必需要再次条件判断,以使只有一个线程进入临界区处理。 对此,转来一段:ui
引用下POSIX的RATIONALE:
Condition Wait Semantics
It is important to note that when pthread_cond_wait() and pthread_cond_timedwait() return without error, the associated predicate may still be false. Similarly, when pthread_cond_timedwait() returns with the timeout error, the associated predicate may be true due to an unavoidable race between the expiration of the timeout and the predicate state change.
The application needs to recheck the predicate on any return because it cannot be sure there is another thread waiting on the thread to handle the signal, and if there is not then the signal is lost. The burden is on the application to check the predicate.
Some implementations, particularly on a multi-processor, may sometimes cause multiple threads to wake up when the condition variable is signaled simultaneously on different processors.
In general, whenever a condition wait returns, the thread has to re-evaluate the predicate associated with the condition wait to determine whether it can safely proceed, should wait again, or should declare a timeout. A return from the wait does not imply that the associated predicate is either true or false.
It is thus recommended that a condition wait be enclosed in the equivalent of a "while loop" that checks the predicate.
从上文能够看出:
1,pthread_cond_signal在多处理器上可能同时唤醒多个线程,当你只能让一个线程处理某个任务时,其它被唤醒的线程就须要继续 wait,while循环的意义就体如今这里了,并且规范要求pthread_cond_signal至少唤醒一个pthread_cond_wait上 的线程,其实有些实现为了简单在单处理器上也会唤醒多个线程.
2,某些应用,如线程池,pthread_cond_broadcast唤醒所有线程,但咱们一般只须要一部分线程去作执行任务,因此其它的线程须要继续wait.因此强烈推荐此处使用while循环.
其实说白了很简单,就是pthread_cond_signal()也可能唤醒多个线程,而若是你同时只容许一个线程访问的话,就必需要使用while来进行条件判断,以保证临界区内只有一个线程在处理。
==============================另外一篇很好的文章===========================
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