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後先等一會兒,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,因爲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註釋掉。
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();
}
}
}
運行結果爲:
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。