AtomicBoolean的用法

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首先先看如下例子
Java代码收藏代码

private static class BarWorker implements Runnable {   private static boolean exists = false;   private String name;   public BarWorker(String name) {    this.name = name;   }   public void run() {    if (!exists) {     exists = true;     System.out.println(name + " enter");     System.out.println(name + " working");     System.out.println(name + " leave");     exists = false;    } else {     System.out.println(name + " give up");    }   }  }

static变量exists用来实现同一时间只有一个worker在工作. 但是假设exists的判断和exists = true;之间有了
其他指令呢
Java代码收藏代码

private static class BarWorker implements Runnable {   private static boolean exists = false;   private String name;   public BarWorker(String name) {    this.name = name;   }   public void run() {    if (!exists) {     try {      TimeUnit.SECONDS.sleep(1);     } catch (InterruptedException e1) {      // do nothing     }     exists = true;     System.out.println(name + " enter");     try {      System.out.println(name + " working");      TimeUnit.SECONDS.sleep(2);     } catch (InterruptedException e) {      // do nothing     }     System.out.println(name + " leave");     exists = false;    } else {     System.out.println(name + " give up");    }   }  }

这时输出是
bar2 enter
bar2 working
bar1 enter
bar1 working
bar1 leave
bar2 leave
看到两个线程同时工作了.
这时可以用AtomicBoolean
Java代码收藏代码

private static class BarWorker implements Runnable {    private static AtomicBoolean exists = new AtomicBoolean(false);    private String name;    public BarWorker(String name) {     this.name = name;    }    public void run() {     if (exists.compareAndSet(false, true)) {      System.out.println(name + " enter");      try {       System.out.println(name + " working");       TimeUnit.SECONDS.sleep(2);      } catch (InterruptedException e) {       // do nothing      }      System.out.println(name + " leave");      exists.set(false);     }else{      System.out.println(name + " give up");     }    }   }

因为它提供了原子性操作,其中exists.compareAndSet(false, true)这个操作把比较和赋值操作组成了一个原子操作,
中间不会提供可乘之机.输出为
bar1 enter
bar1 working
bar2 give up

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