使用jstack定位程序许久没有反应死锁问题

引言

这个情况现象的程序运行的时候迟迟没有输出，我们可以怀疑的死锁的问题，但是怎么去定位这个问题呢，我们还是借助jstack来做。

jstack 163746

这个查看没有太复杂的流程，直接查看堆栈信息最后的一部分就ok

Found one Java-level deadlock:
=============================
"Thread_02":
  waiting to lock monitor 0x00002b578c002178 (object 0x000000058015e5c0, a O1),
  which is held by "Thread_01"
"Thread_01":
  waiting to lock monitor 0x00002b578c006218 (object 0x000000058015f148, a O2),
  which is held by "Thread_02"

Java stack information for the threads listed above:
===================================================
"Thread_02":
        at Demo1_3.lambda$main$1(Demo1_3.java:28)
        - waiting to lock <0x000000058015e5c0> (a O1)
        - locked <0x000000058015f148> (a O2)
        at Demo1_3$$Lambda$2/1418481495.run(Unknown Source)
        at java.lang.Thread.run(Thread.java:745)
"Thread_01":
        at Demo1_3.lambda$main$0(Demo1_3.java:17)
        - waiting to lock <0x000000058015f148> (a O2)
        - locked <0x000000058015e5c0> (a O1)
        at Demo1_3$$Lambda$1/834600351.run(Unknown Source)
        at java.lang.Thread.run(Thread.java:745)

Found 1 deadlock.

信息会很明确告诉有一个死锁，并且指明了代码行17行和28行，定位到这里就方便了，我们拿出源代码看看：

class O1{}
class O2{}
public class Demo1_3 {
    static O1 o1=new O1();
    static  O2 o2=new O2();
    public static void main(String[] args) throws InterruptedException {
        new Thread(()->{
            synchronized (o1){
                    try {
                        Thread.sleep(2000);
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                synchronized (o2){
                    System.out.println("我是线程1，我获得了o1和o2");
                }
            }
        },"Thread_01").start();
        Thread.sleep(1000);
        
        new Thread(()->{
            synchronized (o2){
                synchronized (o1){
                    System.out.println("我是线程2,我获得了o1和o2");
                }
            }
        },"Thread_02").start();
    }
}

可以看到，线程1和2分别取锁定o1和o2，线程1在持有了o1的同时又准备去锁住o2，而线程2则先持有o2再去锁定o1,双方都没有释放，进而死锁了。