0


有谁可以告诉我如何在函数“getDirectHair()”中编码返回值?
我想创建一个类似于 Human 类中的快捷方式的方法,以直接返回 Hair 类的良好类型(下面命名为“h.getDirectHair()”),而不是使用“h.getPerson().getHair()”。
我想使用在 Human 类中声明的 Person<?> 的类型 <?>。

package test;
public class Test {
    public static void main(String[] args) {
        Human<Bob> h = new Human<Bob>();
        Blond blond = h.getPerson().getHair(); // no cast needed, because Human<Bob> is blond
        //how to do if I want to use directly this :
        blond = (!!!) h.getDirectHair();  //need cast !! Blond or Brown ?
    }
}
class Human<T extends Person<?>>{
    private T person = null;
    public T getPerson() {
        return person;
    }
    public /* <?> */ Object getDirectHair(){
        // => I want to return the type <?> of Person<?> 
        // instead of Object, how to ??
        return person.getHair();
    }
}
class Person<T extends Hair> {
    T hair;
    public Person(T hairr) {
        hair = hairr;
    }
    public T getHair() {
        return hair;
    }
}
class Bob extends Person<Blond> {
    public Bob(Blond bean) {
        super(bean);
    }
    public Blond getHair() {
        return super.getHair();
    }
}
class Barack extends Person<Brown> {
    public Barack(Brown bean) {
        super(bean);
    }
    public Brown getHair() {
        return super.getHair();
    }
}
class Hair {
}
class Blond extends Hair {
}
class Brown extends Hair {
}


非常感谢和最好的问候,
大卫。

4

1 回答 1

4

以下是解决问题的方法:

public class Test {
    public static void main(String[] args) {
        Human<Blond, Bob> h = new Human<Blond, Bob>();
        Blond blond = h.getPerson().getHair(); 
        blond = h.getDirectHair();
    }
}

class Human<H extends Hair, T extends Person<H>>{
    private T person = null;
    public T getPerson() {
        return person;
    }
    public H getDirectHair(){
        return person.getHair();
    }
}
于 2013-07-17T11:08:34.597 回答