问题描述
我正尝试使用信号灯严格依次启动10个线程。 也就是说,在执行线程0之后,应该执行线程1,而不是线程2。
但是问题在于,线程是按乱序到达semaphore.acquire()
方法的,因此线程的执行是乱序的。
如何使用信号量但不使用thread.join()
来解决此问题?
public class Main {
private Semaphore semaphore = new Semaphore(1,true);
public static void main(String[] args) {
new Main().start();
}
private void start() {
for (int i = 0; i < 10; i++) {
Runnable runnable = () -> {
try {
semaphore.acquire();
} catch (InterruptedException e) {
e.printstacktrace();
}
System.out.println("In run method " + Thread.currentThread().getName());
semaphore.release();
};
Thread thread = new Thread(runnable);
thread.start();
}
}
}
输出:
In run method Thread-0
In run method Thread-1
In run method Thread-4
In run method Thread-5
In run method Thread-3
In run method Thread-2
In run method Thread-6
In run method Thread-7
In run method Thread-9
In run method Thread-8
解决方法
您需要具有某种排序概念的同步对象。如果您熟悉美国的杂货店,请考虑在熟食店的柜台上“取个号码”,告诉您轮到谁了。
粗略的代码:
class SyncThing {
int turn = 0;
synchronized void waitForTurn(int me) {
while (turn != me)
wait();
}
synchronized void nextTurn() {
turn++;
notifyAll();
}
}
然后声明SyncThing syncThing = new SyncThing();
并因此运行第i个线程:
Runnable runnable = () -> {
syncThing.waitForTurn(i);
System.out.println("In run method " + Thread.currentThread().getName());
syncThing.nextTurn();
};
这是我脑海中输入的,没有提供完整的代码,但应该可以显示出方法。
,private void start() {
final AtomicInteger counter = new AtomicInteger();
for (int i = 0; i < 10; i++) {
final int num = i;
new Thread(() -> {
while (counter.get() != num) {
}
System.out.println("In run method " + Thread.currentThread().getName());
counter.incrementAndGet();
}).start();
}
}