使用Java中的组合和接口进行设计

Question

我为一个问题设计了以下内容:

class Animal {
 // ...
}
class Guppy extends Animal { ... }
class Pigeon extends Animal { ... }

class TailedAnimal extends Animal {
 // ...
}
class Dog extends TailedAnimal { ... }
class Cat extends TailedAnimal { ... } 

class HornedAnimal extends Animal {
 // ...
}
class Ram extends HornedAnimal { ... }

public static void main(String[] args) {
 Animal a = getSomeAnimal();
 a.doSomething();
 if (a instanceof TailedAnimal) {
  // do something
 }
 if (a instanceof HornedAnimal) {
  // do something else
 }
}

Animal,HornedAnimal和TailedAnimal主要用作数据模型.

由于Java不支持多重继承,因此我无法创建Rhinoceros,它是一种角和尾动物.在询问后,有人建议使用组合和界面.我想出了以下内容:

class Animal {
 // ...
}
class Guppy extends Animal { ... }
class Pigeon extends Animal { ... }
class Ram extends Animal implements IHorned { ... }
class Cat extends Animal implements ITailed { ... } 
class Dog extends Animal implements ITailed {
 BasicTail t = new BasicTail();
 public Object getTail() {
  return t.getTail();
 }
 public void setTail(Object in) {
  t.setTail(in);
 }
}

interface ITailed {
 public Object getTail();
 public void setTail(Object in);
 //...
}

class BasicTail implements ITailed {
 Object myTail;
 public Object getTail() { return myTail; }
 public void setTail(Object t) { myTail = t; }
}

interface IHorned {
 // getters and setters
}

public static void main(String[] args) {
 Animal a = getSomeAnimal();
 a.doSomething();
    // how do I check if a is horned or tailed?
}

我的界面有getter和setter.有什么方法可以避免这种情况吗？假设目前没有办法抽象Tails和Horns的行为,并且它们主要用作数据持有者.如何确定我的动物是有角还是长尾？

Answer 1

我在这里建议策略模式.简而言之:

interface TailedAnimal {
    void moveTail();
}
interface HornedAnimal {
    void hitWithHorn();
}
class Rhinoceros() implements TailedAnimal, HornedAnimal {
    private TailedAnimal tail;  //Instantiate it somehow e.g. constructor, setter
    private HornedAnimal horn;  //Instantiate it somehow e.g. constructor, setter
    public void moveTail() {
        tail.moveTail();
    }
    public void hitWithHorn() {
        horn.hitWithHorn();
    }
}

通过使用它,您可以在接口的具体实现中封装行为,并且可以轻松地为少数动物分享完全相同的行为,以及在运行时更改它.

Answer 2

我认为一般来说你必须避免设置器。如果可以，请使用不可变对象，并将其私有数据初始化到其构造函数中。

为了区分动物，我使用了另一种模式，即访客模式。它很冗长，但您不必直接测试您正在处理的动物。

public class Animals {
private Animals() {
}

interface Animal {
    void accept(final AnimalProcessor visitor);
}

interface AnimalProcessor {
    void visitTailed(final TailedAnimal tailedAnimal);

    void visitHorned(final HornedAnimal hornedAnimal);
}

interface TailedAnimal extends Animal {
    void moveTail();
}

interface HornedAnimal extends Animal {
    void hitWithHorns();
}

static class Dog implements TailedAnimal {
    public void moveTail() {
        //To change body of implemented methods use File | Settings | File Templates.
    }

    public void accept(final AnimalProcessor visitor) {
        visitor.visitTailed(this);
    }
}

static class Cat implements TailedAnimal {
    public void moveTail() {
        //To change body of implemented methods use File | Settings | File Templates.
    }

    public void accept(final AnimalProcessor visitor) {
        visitor.visitTailed(this);
    }
}

static class Ram implements HornedAnimal {
    public void hitWithHorns() {
        //To change body of implemented methods use File | Settings | File Templates.
    }

    public void accept(final AnimalProcessor visitor) {
        visitor.visitHorned(this);
    }
}

static class Rhinoceros implements HornedAnimal, TailedAnimal {
    public void hitWithHorns() {
        //To change body of implemented methods use File | Settings | File Templates.
    }

    public void moveTail() {
        //To change body of implemented methods use File | Settings | File Templates.
    }

    public void accept(final AnimalProcessor visitor) {
        visitor.visitTailed(this);
        visitor.visitHorned(this);
    }
}

public static void main(String[] args) {
    Collection<Animal> animals = new ArrayList<Animal>(Arrays.asList(new Dog(), new Cat(), new Rhinoceros()));
    for (final Animal animal : animals) {
        animal.accept(new AnimalProcessor() {
            public void visitTailed(final TailedAnimal tailedAnimal) {
                // you do what you want when it's a tailed animal
            }

            public void visitHorned(final HornedAnimal hornedAnimal) {
                // you do what you want when it's a horned animal
            }
        });
    }
}
}