Java 以特定顺序实现多线程完成的同步改进
问题陈述:巴士可以按任何顺序启动,但会按特定顺序到达目的地。所以,若之前预期的巴士还并没有到达,巴士将等待 在遇到多个丢失的信号并解决它们之后,我最终提出了以下解决方案。 我的问题:首先,我们如何修改下面的代码,使其从设计的角度变得健壮?第二,JAVA中有没有其他的同步机制可以用来解决这个问题Java 以特定顺序实现多线程完成的同步改进,java,multithreading,locking,wait,notify,Java,Multithreading,Locking,Wait,Notify,问题陈述:巴士可以按任何顺序启动,但会按特定顺序到达目的地。所以,若之前预期的巴士还并没有到达,巴士将等待 在遇到多个丢失的信号并解决它们之后,我最终提出了以下解决方案。 我的问题:首先,我们如何修改下面的代码,使其从设计的角度变得健壮?第二,JAVA中有没有其他的同步机制可以用来解决这个问题 public class BusDestination implements Runnable { private boolean[] busArrivalStatus; private List<
public class BusDestination implements Runnable {
private boolean[] busArrivalStatus;
private List<String> busSeq = new ArrayList<>();
private Lock lock = new ReentrantLock();
private Condition prevBusFlag = lock.newCondition();
private CountDownLatch latch;
boolean signalled = false;
public BusDestination(CountDownLatch latch) {
busSeq.add("B1");
busSeq.add("B2");
busSeq.add("B3");
busSeq.add("B4");
busSeq.add("B5");
busArrivalStatus = new boolean[busSeq.size()];
for (int i = 0; i < this.busSeq.size(); i++) {
busArrivalStatus[i] = false;
}
this.latch = latch;
}
public void run() {
String busName = Thread.currentThread().getName();
try{
lock.lock();
int busArrrivalSeq = busSeq.indexOf(busName);
if(busArrrivalSeq==0){
//first bus arrived
System.out.println("******** Bus arriaved : "+busName);
busArrivalStatus[0] = true;
System.out.println(busName+" will signall");
signalled = true;
prevBusFlag.signalAll();
} else {
while(!isValidSeq(busName) && !signalled ){
System.out.println(busName+" going to wait.");
prevBusFlag.await();
}
System.out.println("******** Bus arriaved : "+busName);
busArrivalStatus[busSeq.indexOf(busName)] = true;
signalled = true;//getPreviousBusStatus(busName);
prevBusFlag.signalAll();
}
} catch(Exception ex){
ex.printStackTrace();
System.out.println("Except--"+busName);
}finally{
lock.unlock();
}
latch.countDown();
}
private boolean isValidSeq(String busName) {
int prevIndex = busSeq.indexOf(busName)-1;
if(!busArrivalStatus[prevIndex]){
signalled = false;
}
return busArrivalStatus[prevIndex];
}
public static void main(String[] args) throws InterruptedException {
CountDownLatch latch = new CountDownLatch(5);
BusDestination destination = new BusDestination(latch);
Thread b1 = new Thread(destination);
Thread b2 = new Thread(destination);
Thread b3 = new Thread(destination);
Thread b4 = new Thread(destination);
Thread b5 = new Thread(destination);
b1.setName("B1");
b2.setName("B2");
b3.setName("B3");
b4.setName("B4");
b5.setName("B5");
b4.start();
b5.start();
b3.start();
b1.start();
b2.start();
latch.await();
}
公共类BusDestination实现可运行{
私有布尔[]总线到达状态;
private List busSeq=new ArrayList();
private Lock=new ReentrantLock();
私有条件prevBusFlag=lock.newCondition();
私人倒计时闩锁;
布尔信号=假;
公共总线目的地(倒计时闩锁){
busSeq.添加(“B1”);
busSeq.添加(“B2”);
busSeq.添加(“B3”);
busSeq.添加(“B4”);
busSeq.添加(“B5”);
busArrivalStatus=新布尔值[busSeq.size()];
对于(int i=0;i
}根据您试图实现的目标,此问题可能会对您有所帮助:但是,该问题中的答案侧重于线程完成工作后的同步,这与您想要的并不完全相同 除此之外,倒计时锁不是多余的吗?如果使用signal/wait同步线程,是否仍需要等待它们同时完成 我的想法是使用一个队列来定义公交车的顺序。线程窥视队列(不移除它),如果轮到它,如果从队列中移除元素并向其他线程发出信号。由于对队列的并发访问仅在关键部分完成,因此可以使用简单的
LinkedList
if/else检查busarrivalseq==0
是非常冗余的,它们实际上做了非常相似的事情。使用队列,它可以简化为一种情况
public class BusDestination implements Runnable {
private Lock lock = new ReentrantLock();
private Condition prevBusFlag = lock.newCondition();
private Queue<String> busSeq;
public BusDestination(Queue<String> busSeq) {
this.busSeq = busSeq;
}
public void run() {
String busName = Thread.currentThread().getName();
try {
lock.lock();
while (!busSeq.peek().equals(busName) ) {
System.out.println(busName + " going to wait.");
prevBusFlag.await();
}
System.out.println("******** Bus arriaved : " + busName);
busSeq.remove();
prevBusFlag.signalAll();
} catch (Exception ex) {
ex.printStackTrace();
System.out.println("Except--" + busName);
} finally {
lock.unlock();
}
}
public static void main(String[] args) throws InterruptedException {
Queue<String> busSeq = new LinkedList<>();
busSeq.add("B1");
busSeq.add("B2");
busSeq.add("B3");
busSeq.add("B4");
busSeq.add("B5");
BusDestination destination = new BusDestination(busSeq);
Thread b1 = new Thread(destination);
Thread b2 = new Thread(destination);
Thread b3 = new Thread(destination);
Thread b4 = new Thread(destination);
Thread b5 = new Thread(destination);
b1.setName("B1");
b2.setName("B2");
b3.setName("B3");
b4.setName("B4");
b5.setName("B5");
b4.start();
b5.start();
b3.start();
b1.start();
b2.start();
}
}
公共类BusDestination实现可运行{
private Lock=new ReentrantLock();
私有条件prevBusFlag=lock.newCondition();
专用队列总线;
公共总线目的地(队列总线顺序){
this.busSeq=busSeq;
}
公开募捐{
字符串busName=Thread.currentThread().getName();
试一试{
lock.lock();
而(!busSeq.peek().equals(busName)){
System.out.println(busName+“正在等待”);
prevBusFlag.wait();
}
System.out.println(“*********总线到达:“+busName”);
busSeq.remove();
prevBusFlag.signalAll();
}捕获(例外情况除外){
例如printStackTrace();
System.out.println(“除了--”+busName);
}最后{
lock.unlock();
}
}
公共静态void main(字符串[]args)引发InterruptedException{
Queue busSeq=新链接列表();
busSeq.添加(“B1”);
busSeq.添加(“B2”);
busSeq.添加(“B3”);
busSeq.添加(“B4”);
busSeq.添加(“B5”);
BusDestination=新的BusDestination(busSeq);
线程b1=新线程(目标);
线程b2=新线程(目标);
线程b3=新线程(目标);
线程b4=新线程(目标);
线程b5=新线程(目标);
b1.设定名称(“b1”);
b2.设定名称(“b2”);
b3.设定名称(“b3”);
b4.设定名称(“b4”);
b5.设定名称(“b5”);
b4.开始();
b5.start();
b3.start();
b1.开始();
b2.start();
}
}
这可以通过只使用CountDownLatch实例而不需要lock实例来简化。您试图解决的问题是,线程n都可以并发执行,但在某一点上,线程t必须等到线程t-1到达某一点。这是我的解决办法
public class BusDestination {
private CountDownLatch pre, next;
public BusDestination(CountDownLatch pre, CountDownLatch next) {
this.pre = pre;
this.next = next;
}
public void run() {
// Do work....
pre.await(); // wait until prior thread{s} are done
// If required, do more work...
pre.countDown(); // notify next set of thread{s} that you are done
// If required, do more work...
}
public static void main(String args[]) {
CountDownLatch first = new CountDownLatch(0); // await will return right away.
CountDownLatch second = new CountDownLatch(1);
// Continue creating CountDownLatch....
CountDownLatch preLast = new CountDownLatch(1);
CountDownLatch last = new CountDownLatch(1); // used to wait by the main thread.
BusDestination firstBus = new BusDestination(first, second);
// create more destinations
BusDestination lastBus = new BusDestination(preLast, last);
// start threads....
last.await(); // Waits until all of the threads complete
}
这个解决方案的一个很好的特性是您可以创建一个pip