CS 61B Lab5

其实我不太确定题目是否是这个意思，我就按照我的理解来答题，欢迎来讨论。

part one    

a b c d 答案都在以下代码中

分析：array的情况和variable情况是一样的，除了代码中array 的cast有点点不一样。。。在compile过程中，array之间相互assign看的是static type，type一致或者是子类assign给了父类或者父类cast成子类之后assign给子类，这三种情况是可以通过compile time的。但是在run time过程中，看的是dynamic type，必须要一致，才能run。所以在d中最后，Dog type 的 twodogs 指向了 Teddy type，如此一来，在最后cast过后，就既能通过compile time，又能run起来了。

package lab5;

public class Dog{
    protected int legs;
    protected String voice;
    protected String name;
    Dog(){
        legs = 4;
        voice = "wang";
        name = "dog";
    }
/**************************************************************/
    public static void main(String[] arg){
        Dog onedog;
        Teddy oneTed = new Teddy();
        onedog = oneTed;
//        oneTed = onedog;    compile error
        oneTed = (Teddy)onedog;
        
        onedog = new Dog();
//        oneTed = (Teddy)onedog;         runtime error
        
        
/**************************************************************/        
        
        Dog[] Twodogs;
        Teddy[] TwoTeddy=new Teddy[2];
        

        TwoTeddy[0] = new Teddy();
        TwoTeddy[1] = new Teddy();
        
        Twodogs = TwoTeddy;
//        TwoTeddy = Twodogs;//compile error;
        TwoTeddy = (Teddy[])Twodogs;
        
        Twodogs = new Dog[2];
//        TwoTeddy = (Teddy[])Twodogs;        runtime error
        
/**************************************************************/
        Twodogs = new Teddy[2];
        Twodogs = TwoTeddy;
//        TwoTeddy = Twodogs;  compile error
        TwoTeddy = (Teddy[])Twodogs; // there is no run error or compile error
                
    }
}

 

Part Two

(a) java will compile the result

package lab5;

public interface Animals {
     public void Shout();
}



package lab5;

public class Dog{
    protected int legs;
    protected String voice;
    protected String name;
    Dog(){
        legs = 4;
        voice = "wang";
        name = "dog";
    }
    public void Shout(){
        System.out.println(voice);
    }
}

（b）NO

（c）NO

  (d) Yes

Part Three

（a）Yes, there is no difference.

  (bc)  the situation is like this below. 

Part Four

(a) call the subclass method

(b) run time error

（c）课上讲过的方法 用super调用superclass method