所以我想要实现的是两个线程轮流执行他们的任务。我原来只有一个问题;
所以我打算做一个代码的小例子,然后我发现即使有锁,我也无法使它工作。所以我的第二个问题是;我如何使代码按预期工作?我看到它的方式,它应该工作,但那只是我:)。
public class App {
Lock lock = new ReentrantLock();
Condition cond1 = lock.newCondition();
Condition cond2 = lock.newCondition();
public App() {
Thread thread1 = new Thread(new Runnable() {
@Override
public void run() {
try {
while (true) {
lock.lock();
System.out.println("Thread 1");
cond2.signalAll();
cond1.await();
lock.unlock();
}
} catch (InterruptedException e) {
}
}
});
thread1.start();
Thread thread2 = new Thread(new Runnable() {
@Override
public void run() {
try {
while (true) {
lock.lock();
cond2.await();
System.out.println(" Thread 2");
cond1.signalAll();
lock.unlock();
}
} catch (InterruptedException e) {
}
}
});
thread2.start();
}
public static void main(String[] args) {
new App();
}
}
答案 0 :(得分:2)
(1)await()
通常用于循环中;不这样做是一个错误的迹象。
while( some condition not met )
cond.await();
(2)unlock()
应位于finally
块
(3)signal()
仅表示当前正在等待的线程;如果没有线程等待,信号就会丢失。
lock.lock();
cond.signal(); // lost
cond.await(); // won't wake up
(4)对于像这样的简单案例,使用好的旧synchronized
没有错。实际上,在使用更多" advanced"之前,你最好先理解它。东西。
(5)解决方案:
Lock lock = new ReentrantLock();
Condition cond = lock.newCondition();
int turn=1; // 1 or 2
// thread1
lock.lock();
try
{
while (true)
{
while(turn!=1)
cond.await();
System.out.println("Thread 1");
turn=2;
cond.signal();
}
}
finally
{
lock.unlock();
}
// thread2
// switch `1` and `2`
(6)一个线程环,每个都唤醒下一个
int N = 9;
Thread[] ring = new Thread[N];
for(int i=0; i<N; i++)
{
final int ii = i+1;
ring[i] = new Thread(()->
{
while(true)
{
LockSupport.park(); // suspend this thread
System.out.printf("%"+ii+"d%n", ii);
LockSupport.unpark(ring[ii%N]); // wake up next thread
// bug: spurious wakeup not handled
}
});
}
for(Thread thread : ring)
thread.start();
LockSupport.unpark(ring[0]); // wake up 1st thread
答案 1 :(得分:0)
您可以使用thread notify()和wait():Oracle Documentation