思路是創建兩個字符串a和b,再創建兩個線程A和B,讓每個線程都用synchronized鎖住字符串(A先鎖a,再去鎖b;B先鎖b,再鎖a),如果A鎖住a,B鎖住b,A就沒辦法鎖住b,B也沒辦法鎖住a,這時就陷入了死鎖。直接貼代碼:
public class DeadLock { public static String obj1 = "obj1"; public static String obj2 = "obj2"; public static void main(String[] args){ Thread a = new Thread(new Lock1()); Thread b = new Thread(new Lock2()); a.start(); b.start(); } } class Lock1 implements Runnable{ @Override public void run(){ try{ System.out.println("Lock1 running"); while(true){ synchronized(DeadLock.obj1){ System.out.println("Lock1 lock obj1"); Thread.sleep(3000);//獲取obj1后先等一會兒,讓Lock2有足夠的時間鎖住obj2 synchronized(DeadLock.obj2){ System.out.println("Lock1 lock obj2"); } } } }catch(Exception e){ e.printStackTrace(); } } } class Lock2 implements Runnable{ @Override public void run(){ try{ System.out.println("Lock2 running"); while(true){ synchronized(DeadLock.obj2){ System.out.println("Lock2 lock obj2"); Thread.sleep(3000); synchronized(DeadLock.obj1){ System.out.println("Lock2 lock obj1"); } } } }catch(Exception e){ e.printStackTrace(); } } }
運行的結果如圖所示:
可以看到,Lock1獲取obj1,Lock2獲取obj2,但是它們都沒有辦法再獲取另外一個obj,因為它們都在等待對方先釋放鎖,這時就是死鎖。
如果我們只運行Lock1呢?修改一下main函數,把線程b注釋掉。
public class DeadLock { public static String obj1 = "obj1"; public static String obj2 = "obj2"; public static void main(String[] args){ Thread a = new Thread(new Lock1()); //Thread b = new Thread(new Lock2()); a.start(); //b.start(); } } class Lock1 implements Runnable{ @Override public void run(){ try{ System.out.println("Lock1 running"); while(true){ synchronized(DeadLock.obj1){ System.out.println("Lock1 lock obj1"); Thread.sleep(3000); synchronized(DeadLock.obj2){ System.out.println("Lock1 lock obj2"); } } } }catch(Exception e){ e.printStackTrace(); } } } class Lock2 implements Runnable{ @Override public void run(){ try{ System.out.println("Lock2 running"); while(true){ synchronized(DeadLock.obj2){ System.out.println("Lock2 lock obj2"); Thread.sleep(3000); synchronized(DeadLock.obj1){ System.out.println("Lock2 lock obj1"); } } } }catch(Exception e){ e.printStackTrace(); } } }
運行結果為:
由於沒有其它線程和Lock1爭奪obj1和obj2,Lock1可以不斷地循環獲取並釋放它們,這時沒有死鎖。