IOTXING

记录技术学习之路

0%

synchorinzed的一个面试题

Posted on Edited on Changyan:

背景

前两天一个朋友面试，遇到了一个题目

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
import java.util.LinkedList;
 
  public class Stack {
    LinkedList list = new LinkedList();
    public synchronized void push(Object x) {
      synchronized(list) {
        list.addLast( x );
        notify();
      }
    }
 
    public synchronized Object pop()
      throws Exception {
      synchronized(list) {
        if( list.size() <= 0 ) {
          wait();
        }
        return list.removeLast();
      }
    }
  }

上面的代码，一般情况下工作正常，问什么时候会有问题，并且问题的根源是什么

现象

既然这么问，那么肯定是有问题的，下面就写代码复现一下

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
public static void main(String[] args)
   {
       Stack stack = new Stack();
       new Thread(() -> {
           System.out.println("push thread" + Thread.currentThread().getName());
           while (true)
           {
               try
               {
                   stack.push("asd");
               } catch (Exception e)
               {
                   System.out.println("sd");
               }    
           }
           
       }).start();
       new Thread(() -> {
           System.out.println("pop thread" + Thread.currentThread().getName());
           while (true)
           {
               try
               {
                   stack.pop();
               } catch (Exception e)
               {
                   e.printStackTrace();
               }
           }     
       }).start();  
   }

果不其然，代码跑起来之后就卡主了。dump了一下进程，发现是出现了死锁

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
"Thread-1" #12 prio=5 os_prio=31 tid=0x00007f8848873000 nid=0x5803 in Object.wait() [0x000070000a55b000]
   java.lang.Thread.State: WAITING (on object monitor)
	at java.lang.Object.wait(Native Method)
	- waiting on <0x000000076ada6460> (a Stack)
	at java.lang.Object.wait(Object.java:502)
	at Stack.pop(Stack.java:32)
	- locked <0x000000076ada6610> (a java.util.LinkedList)
	- locked <0x000000076ada6460> (a Stack)
	at Stack.lambda$main$1(Stack.java:69)
	at Stack$$Lambda$2/381259350.run(Unknown Source)
	at java.lang.Thread.run(Thread.java:748)

"Thread-0" #11 prio=5 os_prio=31 tid=0x00007f884701c800 nid=0x5603 waiting for monitor entry [0x000070000a458000]
   java.lang.Thread.State: BLOCKED (on object monitor)
	at Stack.push(Stack.java:14)
	- waiting to lock <0x000000076ada6610> (a java.util.LinkedList)
	- locked <0x000000076ada6460> (a Stack)
	at Stack.lambda$main$0(Stack.java:53)
	at Stack$$Lambda$1/1607521710.run(Unknown Source)
	at java.lang.Thread.run(Thread.java:748)

pop线程持有list的锁，等待获取到stack的锁，而push线程持有stack的锁，等待获取list的锁，因此就出现了死锁导致卡主。

而这种情况的产生，是因为在代码中调用了wait()，因为没有指明对象，所以默认的就是对当前对象执行wait()。因为我们在pop方法中用了两个synchronized，其实就是拿了两个锁，一个是当前对象的锁，一个是list的锁。直接调用wait，释放了this对象的锁，然后push线程拿到了，但是因为pop线程调用了wait之后挂起来了，所以list的锁也就没有释放掉。现在就属于pop等待this对象的锁，而push持有this对象，等待list的锁，因此死锁就产生了。
如果把wait改成list.wait()，会是下面的清空

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
"Thread-1" #15 prio=5 os_prio=0 cpu=31.25ms elapsed=4.21s tid=0x0000019b436f9000 nid=0x1a5c in Object.wait()  [0x0000005c5b0fe000]
   java.lang.Thread.State: WAITING (on object monitor)
	at java.lang.Object.wait(java.base@13.0.2/Native Method)
	- waiting on <0x00000006206104b0> (a java.util.LinkedList)
	at java.lang.Object.wait(java.base@13.0.2/Object.java:326)
	at Stack.pop(Stack.java:19)
	- locked <0x00000006206104b0> (a java.util.LinkedList)
	- locked <0x0000000620610240> (a Stack)
	at Stack.lambda$main$1(Stack.java:36)
	at Stack$$Lambda$16/0x0000000800ba5040.run(Unknown Source)
	at java.lang.Thread.run(java.base@13.0.2/Thread.java:830)
	
"Thread-0" #14 prio=5 os_prio=0 cpu=0.00ms elapsed=4.21s tid=0x0000019b436f7000 nid=0x453c waiting for monitor entry  [0x0000005c5affe000]
   java.lang.Thread.State: BLOCKED (on object monitor)
	at Stack.push(Stack.java:8)
	- waiting to lock <0x0000000620610240> (a Stack)
	at Stack.lambda$main$0(Stack.java:29)
	at Stack$$Lambda$15/0x0000000800ba4440.run(Unknown Source)
	at java.lang.Thread.run(java.base@13.0.2/Thread.java:830)

情况刚好相反，pop等待list的锁，push等待stack的锁。
因此如果想要避免这种情况，需要把pop函数上面的synchronized去掉，只保留一个锁即可