Java:在超类方法上同步

Question

abstract class Basic (){
    public synchronized void basicMethod(String string){
        //Some actions here
    }
}

public class A extends Basic{
    public void aMethod(){
        //Some actions here
    }
}

public class B extends Basic{
    public void bMethod(){
        //Some actions here
    }
}

Basic a = new A();
Basic b = new B();

a.basicMethod(); // acquires lock
b.basicMethod(); //Same lock?

换句话说 - 锁也与具体的Object或Super类有关吗？

Answer 1

不同的锁.甲synchronized实例方法同步与该特定对象相关联的监视器上的实例,因此每个的a和b具有其自己的显示器.

如果你有

abstract class Basic (){
    public synchronized void basicMethod(String string){
        //Some actions here
    }
}

public class A extends Basic{
    public synchronized void aMethod(){
        //Some actions here
    }
}

然后调用a.basicMethod()并a.aMethod()锁定同一个监视器.