Question

我正在尝试使用多个线程中的共享整数对象按顺序打印从1到10的数字。当使用共享对象作为AtomicInteger时，程序可以正常工作，但是当使用普通的Integer对象时，程序会引发异常，我不知道为什么会发生这种情况。

使用AtomicInteger的程序

import java.util.concurrent.atomic.AtomicInteger;  

public class ThreadingProblem {

  public static void main(String[] args) {  
    AtomicInteger sharedInt = new AtomicInteger(0);  
    Thread t1 = new Thread(new ThreadingPrintingTask(sharedInt), "PrinterThread");  
    Thread t2 = new Thread(new ThreadingIncrementingTask(sharedInt), "IncrementerThread");  
    t1.start();  
    t2.start();  
  }  
}  

class ThreadingPrintingTask implements Runnable {    

  private AtomicInteger sharedObject;  

  public ThreadingPrintingTask(AtomicInteger sharedObject) {  
    this.sharedObject = sharedObject;  
  }  

  @Override  
  public void run() {  
    try {  
      synchronized (sharedObject) {  
        while (true) {  
          sharedObject.wait();  
          System.out.println("Shared object value is: " + sharedObject);  
          sharedObject.notify();  
        }  
      }  
    }  
    catch (InterruptedException e) {  
    }  

  }  
}  

class ThreadingIncrementingTask implements Runnable {  

  private AtomicInteger sharedObject;  

  public ThreadingIncrementingTask(AtomicInteger sharedObject) {  
    this.sharedObject = sharedObject;  
  }  

  @Override  
  public void run() {  
    synchronized (sharedObject) {  
      while (this.sharedObject.get() < 10) {  
        this.sharedObject.incrementAndGet();  
        this.sharedObject.notify();  
        try {  
          this.sharedObject.wait();  
        }  
        catch (InterruptedException e) {  
          // TODO Auto-generated catch block  
          e.printStackTrace();  
        }  
      }  
    }  
  }  
}

输出

Shared object value is: 1
Shared object value is: 2
Shared object value is: 3
Shared object value is: 4
Shared object value is: 5
Shared object value is: 6
Shared object value is: 7
Shared object value is: 8
Shared object value is: 9
Shared object value is: 10

带有普通整数对象的程序

public class ThreadingProblem {

  public static void main(String[] args) {
    Integer sharedInt = new Integer(0);
    Thread t1 = new Thread(new ThreadingPrintingTask(sharedInt), "PrinterThread");
    Thread t2 = new Thread(new ThreadingIncrementingTask(sharedInt), "IncrementerThread");
    t1.start();
    t2.start();
  }
}

class ThreadingPrintingTask implements Runnable {

  private Integer sharedObject;

  public ThreadingPrintingTask(Integer sharedObject) {
    this.sharedObject = sharedObject;
  }

  @Override
  public void run() {
    try {
      synchronized (sharedObject) {
        while (true) {
          sharedObject.wait();
          System.out.println("Shared object value is: " + sharedObject);
          sharedObject.notify();
        }
      }
    }
    catch (InterruptedException e) {
    }

  }
}

class ThreadingIncrementingTask implements Runnable {

  private Integer sharedObject;

  public ThreadingIncrementingTask(Integer sharedObject) {
    this.sharedObject = sharedObject;
  }

  @Override
  public void run() {
    synchronized (sharedObject) {
      while (this.sharedObject < 10) {
        this.sharedObject++;
        this.sharedObject.notify();
        try {
          this.sharedObject.wait();
        }
        catch (InterruptedException e) {
          // TODO Auto-generated catch block
          e.printStackTrace();
        }
      }
    }
  }
}

输出

Exception in thread "IncrementerThread" java.lang.IllegalMonitorStateException
    at java.lang.Object.notify(Native Method)
    at com.itiviti.apps.catalys.shared.mock.ThreadingIncrementingTask.run(ThreadingProblem.java:52)
    at java.lang.Thread.run(Unknown Source)

Answer 1

this.sharedObject++;并没有您想象的那样。

由于Integer是不可变的，因此无法更改现有共享对象。相反，此操作要做的是将值装箱到int中，将其递增，然后将其装箱到另一个Integer实例中。

因此，您的代码（几乎*）等效于以下内容：

int temp = this.sharedObject.intValue();
temp = temp + 1;
this.sharedObject = new Integer(temp);

由于此时您的对象不再是同一实例，因此您的synchronized块将不会与wait() / notify()调用对齐。

请注意，这与AtomicInteger的原子性无关，只是与++运算符在Integer上的工作方式有关。

*实际上，您可能会获得一个缓存的实例而不是new Integer()，但是它仍然是一个不同的实例，因为它代表了一个不同的int值。

Java等待通知共享整数对象

1 个答案: