public class Deadlock implements Runnable {
public int flag = 1;
static Object o1 = new Object(); //去掉任何一个静态则锁不住,why?
static Object o2 = new Object();
public void run() {
System.out.println("flag = "+flag);
if(flag == 1) {
synchronized(o1) {
try {Thread.sleep(500); } catch (InterruptedException e) {e.printStackTrace();}
synchronized(o2) {System.out.println("1");}}}
if(flag == 0) {
synchronized(o2) {
try {Thread.sleep(500);} catch (InterruptedException e)
{e.printStackTrace();}
synchronized(o1) {System.out.println("0");}}}
}
public static void main(String[] args) {
Deadlock d1 = new Deadlock();
Deadlock d2 = new Deadlock();
d1.flag = 1;
d2.flag = 0;
Thread t1 = new Thread(d1);
Thread t2 = new Thread(d2);
t1.start();
t2.start();
}
}
public int flag = 1;
static Object o1 = new Object(); //去掉任何一个静态则锁不住,why?
static Object o2 = new Object();
public void run() {
System.out.println("flag = "+flag);
if(flag == 1) {
synchronized(o1) {
try {Thread.sleep(500); } catch (InterruptedException e) {e.printStackTrace();}
synchronized(o2) {System.out.println("1");}}}
if(flag == 0) {
synchronized(o2) {
try {Thread.sleep(500);} catch (InterruptedException e)
{e.printStackTrace();}
synchronized(o1) {System.out.println("0");}}}
}
public static void main(String[] args) {
Deadlock d1 = new Deadlock();
Deadlock d2 = new Deadlock();
d1.flag = 1;
d2.flag = 0;
Thread t1 = new Thread(d1);
Thread t2 = new Thread(d2);
t1.start();
t2.start();
}
}
如果属性不声明为静态的。 那么你实例化出来Deadlock中的 o1 o2 都是不同的对象。 你的锁就当然没用了。
属性声明为静态后 。不管你实例化多少个Deadlock对象,每个Deadlock对象里面的 o1 o2 都是同一个对象