lotusswan

為了運(yùn)用多態(tài)性，當(dāng)編寫子類時(shí)，你肯定會(huì)覆寫超類的方法。當(dāng)你覆寫超類的方法時(shí)，你是否想到過更改方法的返回類型。如果你沒有，那祝賀你，你不曾為此煩惱過；如果你想到過，那你肯定是非常的郁悶。在Java 5.0以前，你是無法更改超類定義的方法的返回類型的。不過當(dāng)你使用Java 5.0以后，你會(huì)驚喜地發(fā)現(xiàn)，此煩惱不復(fù)存在了。

下面的示例顯示了這一新特性的用法：

package ?com.jiang.tiger.chap1;

class ?Point2D? {
?????? protected ? int ?x,?y;

?????? public ?Point2D(?)? {
???????? this .x = 0 ;
???????? this .y = 0 ;
??????}
??
?????? public ?Point2D( int ?x,? int ?y)? {
???????? this .x? = ?x;
???????? this .y? = ?y;
??????}
??????
?????? public ?String?toString()? {
?????????? return ? " the?position?is?x?=? " ? + ?x? + ? " ,y?=? " ? + ?y?;
??????}
????}

???? class ?Point3D? extends ?Point2D? {
?????? protected ? int ?z;

?????? public ?Point3D( int ?x,? int ?y)? {
???????? this (x,?y,? 0 );
??????}
?
?????? public ?Point3D( int ?x,? int ?y,? int ?z)? {
???????? this .x? = ?x;
???????? this .y? = ?y;
???????? this .z? = ?z;?
??????}
??????
?????? public ?String?toString()? {
?????????? return ? " the?position?is?x?=? " ? + ?x? + ? " ,y?=? " ? + ?y? + ? " ,z=? " ? + ?z;
??????}
????}

???? class ?Position2D? {
??????Point2D?location;
?
?????? public ?Position2D(?)? {
???????? this .location? = ? new ?Point2D(?);
??????}

?????? public ?Position2D( int ?x,? int ?y)? {
???????? this .location? = ? new ?Point2D(x,?y);
??????}

?????? public ?Point2D?getLocation(?)? {
???????? return ?location;
??????}
????}

???? public ? class ?Position3D? extends ?Position2D? {
??????Point3D?location;
?
?????? public ?Position3D( int ?x,? int ?y,? int ?z)? {
???????? this .location? = ? new ?Point3D(x,?y,?z);
??????}

?????? public ?Point3D?getLocation(?)? {
???????? return ?location;
??????}
????????????
?????? public ? static ? void ?main(String[]?args)? {
??????????Position2D?position? = ? new ?Position3D( 12 ,? 23 ,? 36 );
??????????Point3D?clone? = ?(Point3D)position.getLocation();
??????????System.out.println(clone);
??????}
????}

為驗(yàn)證功能是否正確，我們看看運(yùn)行結(jié)果：

the position is x = 12,y = 23,z= 36