CS 61B Lab5
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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
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