Java的抽象方法提高我重寫

Question

假設我們有這個類

Animal.java

public class Animal { 

    private String name; 

    public Animal(String name) { 
     this.name = name; 
    } 

    public void shout() { 
     /* must be override */ 
    } 

    public String getName() { 
     return name; 
    } 
} 

Driver.java

import java.util.ArrayList; 
import java.util.List; 

public class Driver { 
    public static void main(String[] args) { 

    List<Animal> animals = new ArrayList<Animal>(); 
    animals.add(new Animal("Dog") { 

     @Override 
     public void shout() { 
      System.out.println(getName() + " sounds like: woof woof woof"); 
     } 
    }); 

    animals.add(new Animal("Cat") { 

     @Override 
     public void shout() { 
      System.out.println(getName() + " sounds like: meow meow meow-ow"); 
     } 
    }); 

    for(Animal a : animals) 
     a.shout(); 
    } 
} 

我怎樣才能提高我實施方法shout（）或其他程序員？假設沒有理由讓動物類抽象化，併爲Dog，Cat等製作類。這擴展了動物

是否有設計模式？

Answer 1

嗯策略模式可以幫助你。

public class Animal { 

    private Shoutable shoutable; 
    private String name; 

    public Animal(String name) { 
     this.name = name; 
    } 

    public void shout() { 
     shoutable.shout(); 
    } 

    public String getName() { 
     return name; 
    } 

    public void setShout(Shoutable shoutable){ 
     this.shoutable=shoutable; 
    } 
} 


public interface Shoutable{ 
    void shout(); 
} 

public class DogShout implements Shoutable{ 
    @Override 
    public void shout{ 
    System.out.println("woff gua guaa gua woff"); 
    } 
} 

Answer 2

您可以在基類中實現shout()並添加返回動物聲音的方法voice()。

像這樣：

public class Animal { 

private String voice; 

private String name; 

public Animal(String name, String voide) { 
    this.name = name; 
    this.voice = voice; 
} 

public void shout() { 
    System.out.println(getName() + " sounds like: "+ getVoice()); 
} 

public String getName() { 
    return name; 
} 

public String getVoice() { return voice } 
} 

在創建動物代碼：

animals.add(new Animal("Dog", "woof")); 

animals.add(new Animal("Cat", "meow"); 

Answer 3

如果你想貓與狗做動物做的事情，而不被動物抽象，就可以了。

public class Cat extends Animal 

，將工作，如果動物是抽象性或正常的，完全實現類。您將使用實現而不是擴展的接口。

Answer 4

你可以做到以下幾點：

public class Animal { 

    private String name; 
    private Shoutable shoutable; 

    public Animal(String name, Shoutable shoutable) { 
     this.name = name; 
     this.shoutable = shoutable; 
    } 

    public void shout() { 
     shoutable.shout(); 
    } 

    public String getName() { 
     return name; 
    } 
} 

的Shoutable接口：

public interface Shoutable { 
    void shout(); 
} 

而且有shoutable實現爲匿名內部類：

import java.util.ArrayList; 
import java.util.List; 

public class Driver { 
    public static void main(String[] args) { 

    List<Animal> animals = new ArrayList<Animal>(); 
    animals.add(new Animal("Dog", new Shoutable() { 
       @Override 
       public void shout() { 
        System.out.println(getName() + " sounds like: woof woof woof"); 
       } 
     }); 

    animals.add(new Animal("Cat", new Shoutable() { 

     @Override 
     public void shout() { 
      System.out.println(getName() + " sounds like: meow meow meow-ow"); 
     } 
    }); 

    for(Animal a : animals) 
     a.shout(); 
    } 
}