package thread;
public class TT2 implements Runnable{
int b = 100;
public synchronized void m1() throws Exception{
b = 1000;
Thread.sleep(5000);
System.out.println("b = " + b);
}
public synchronized void m2() throws Exception{
Thread.sleep(7000);
b = 2000;
}
public void run() {
try {
m1();
}catch(Exception e) {
e.printStackTrace();
}
}
public static void main(String[] args)throws Exception{
TT2 tt = new TT2();
Thread t = new Thread(tt);
t.start();
tt.m2();
System.out.println(tt.b);
}
}
打印的结果是: 1000 b=1000
我不明白这个结果是怎么来的,为什么不是先执行m1的锁呢