CyclicBarrier原理分析 - 969251639/study GitHub Wiki
CyclicBarrier可以让一组线程卡在某一个点,直到CyclicBarrier内部的所有线程都执行到该点才继续运行下去
它有两个构造方法
public CyclicBarrier(int parties, Runnable barrierAction) {
if (parties <= 0) throw new IllegalArgumentException();//参与的线程数量小于0,抛出异常
this.parties = parties;//参与的线程数量
this.count = parties;//参与的线程数量
this.barrierCommand = barrierAction;//钩子,回调方法
}
public CyclicBarrier(int parties) {//参与的线程数量
this(parties, null);
}
它的主要成员变量如下:
/** 控制同步状态的锁 */
private final ReentrantLock lock = new ReentrantLock();
/** 控制线程是否继续执行的条件队列 */
private final Condition trip = lock.newCondition();
/** 参与的线程数 */
private final int parties;
/* 当可以继续执行时的回调方法 */
private final Runnable barrierCommand;
/** 重置用 */
private Generation generation = new Generation();
/** 记录还有多少线程在等待,即有多少个线程先执行完了的数量 */
private int count;
下面开始分析等待执行操作,有两个一带有超时,一个没有
public int await() throws InterruptedException, BrokenBarrierException {
try {
return dowait(false, 0L);
} catch (TimeoutException toe) {
throw new Error(toe); // cannot happen
}
}
public int await(long timeout, TimeUnit unit)
throws InterruptedException,
BrokenBarrierException,
TimeoutException {
return dowait(true, unit.toNanos(timeout));
}
不管哪种方式最后都是调用dowait方法
private int dowait(boolean timed, long nanos)
throws InterruptedException, BrokenBarrierException,
TimeoutException {
final ReentrantLock lock = this.lock;//获取锁
lock.lock();//上锁
try {
final Generation g = generation;
if (g.broken)//是否有被外界打破
throw new BrokenBarrierException();
if (Thread.interrupted()) {//线程被中断
breakBarrier();//重置
throw new InterruptedException();//抛出中断异常
}
int index = --count;//获取等待线程的数量
if (index == 0) { // tripped 如果等待线程的数量为0,即这个if是最后一个线程执行的分支
boolean ranAction = false;//回调执行是否成功的标记
try {
final Runnable command = barrierCommand;
if (command != null)//回调不为空
command.run();//调用回调方法
ranAction = true;//标记回调方法执行成功
nextGeneration();//继续生成下一次门栓给generation,里面会唤醒下面的所有线程从await中返回,也就是这一个方法会触发所有线程继续往下走
return 0;//返回等待线程的数量
} finally {
if (!ranAction)//如果回调不成功
breakBarrier();//重置门栓
}
}
// loop until tripped, broken, interrupted, or timed out
for (;;) {//执行到这里说明不是最后线程进入的分支
try {
if (!timed)//没有超时的控制
trip.await();//等待
else if (nanos > 0L)//超时控制
nanos = trip.awaitNanos(nanos);/等待nanos纳秒
} catch (InterruptedException ie) {
if (g == generation && ! g.broken) {
breakBarrier();
throw ie;
} else {
// We're about to finish waiting even if we had not
// been interrupted, so this interrupt is deemed to
// "belong" to subsequent execution.
Thread.currentThread().interrupt();
}
}
if (g.broken)//如果在等待被唤醒后发现门栓被外界打破
throw new BrokenBarrierException();
if (g != generation)//这里的相等意味着要么超时,要么最后一个线程没有执行到nextGeneration();也就是执行回调的时候抛异常了,那么继续等待,但在回调异常那里又会唤醒这些线程,最后进入到上面的那个if判断,抛出BrokenBarrierException异常,否则都从这里退出
return index;
if (timed && nanos <= 0L) {//超时
breakBarrier();//重置
throw new TimeoutException();
}
}
} finally {//解锁
lock.unlock();
}
}
打破门栓
private void breakBarrier() {
generation.broken = true;//设置打破标记为true
count = parties;//重置
trip.signalAll();//唤醒等待的所有线程
}
最后一个线程正常执行的时候还会生成下一个generation,触发所有等待的线程继续往下走
private void nextGeneration() {
// signal completion of last generation
trip.signalAll();//唤醒所有线程
// set up next generation
count = parties;//重置
generation = new Generation();//生成下一个Generation
}
private static class Generation {
boolean broken = false;
}
另外还有一个重置方法也是很简单
public void reset() {
final ReentrantLock lock = this.lock;
lock.lock();
try {
breakBarrier(); // break the current generation
nextGeneration(); // start a new generation
} finally {
lock.unlock();
}
}