Java多线程之死锁的出现和解决方法

//首先我们先定义两个final的对象锁.可以看做是共有的资源.
 final Object lockA = new Object();
 final Object lockB = new Object();
//生产者A
 class ProductThreadA implements Runnable{
   @Override
   public void run() {
//这里一定要让线程睡一会儿来模拟处理数据 ,要不然的话死锁的现象不会那么的明显.这里就是同步语句块里面,首先获得对象锁lockA,然后执行一些代码,随后我们需要对象锁lockB去执行另外一些代码.
     synchronized (lockA){
     //这里一个log日志
       Log.e("CHAO","ThreadA lock lockA");
       try {
         Thread.sleep(2000);
       } catch (InterruptedException e) {
         e.printStackTrace();
       }
       synchronized (lockB){
        //这里一个log日志
         Log.e("CHAO","ThreadA lock lockB");
         try {
           Thread.sleep(2000);
         } catch (InterruptedException e) {
           e.printStackTrace();
         }

       }
     }
   }
 }
 //生产者B
 class ProductThreadB implements Runnable{
 //我们生产的顺序真好好生产者A相反,我们首先需要对象锁lockB,然后需要对象锁lockA.
   @Override
   public void run() {
     synchronized (lockB){
      //这里一个log日志
       Log.e("CHAO","ThreadB lock lockB");
       try {
         Thread.sleep(2000);
       } catch (InterruptedException e) {
         e.printStackTrace();
       }
       synchronized (lockA){
        //这里一个log日志
         Log.e("CHAO","ThreadB lock lockA");
         try {
           Thread.sleep(2000);
         } catch (InterruptedException e) {
           e.printStackTrace();
         }

       }
     }
   }
 }
 //这里运行线程
ProductThreadA productThreadA = new ProductThreadA();
ProductThreadB productThreadB = new ProductThreadB();

   Thread threadA = new Thread(productThreadA);
   Thread threadB = new Thread(productThreadB);
   threadA.start();
   threadB.start();

10-20 14:54:39.940 18162-18178/? E/CHAO: ThreadA lock lockA
10-20 14:54:39.940 18162-18179/? E/CHAO: ThreadB lock lockB