问题描述

菜鸟饼 能不能用java的工厂模式处理下面的问题。 求代码 和 注释

 interface Animal{
    public void say();
}

class Mammal implements Animal{
    public void say(){
        System.out.println("哺乳动物");
    }
}

class  Reptiles implements Animal{
    public void say(){
        System.out.println("爬行动物");
    }
}

class Whale implements Animal{
    private Animal animal;
    public Whale(Animal animal){
        this.animal = animal;
    }
    public void say(){
        System.out.println("鲸喷水");
        animal.say();
    }
}

class Dog implements Animal{
    private Animal animal;
    public Dog(Animal animal){
        this.animal = animal;
    }
    public void say(){
        System.out.println("狗汪汪叫");
        animal.say();
    }
}

class Lizards implements Animal{
    private Animal animal;
    public Lizards(Animal animal){
        this.animal = animal;
    }
    public void say(){
        System.out.println("蜥蜴四角爬行");
        animal.say();
    }
}

class Snake implements Animal{
    private Animal animal;
    public Snake(Animal animal){
        this.animal = animal;
    }
    public void say(){
        System.out.println("蛇有毒");
    }
}

class Gecko implements Animal{
    private Animal lizards;
    public Gecko(Animal lizards){
        this.lizards = lizards;
    }
    public void say(){
        System.out.println("壁虎是蜥蜴的一种，它比较小");
        lizards.say();
    }
}

class MonitorLizard implements Animal{
    private Animal animal;
    public MonitorLizard(Animal animal){
        this.animal = animal;
    }
    public void say(){
        System.out.println("巨蜥是蜥蜴的一种，它比较大");
        animal.say();
    }
}

class Test{
    public static void main(String args []){
        int flag = 4;
        if(flag == 1){
            Mammal mammal = new Mammal();
            Whale whale = new Whale(mammal);
            whale.say();
        }
        else if(flag == 2){
            Reptiles reptiles = new Reptiles();
            Lizards lizards = new Lizards(reptiles);
            lizards.say();
        }
        else if(flag == 3){
            Reptiles reptiles1 = new Reptiles();
            Lizards lizards1 = new Lizards(reptiles1);
            Gecko gecko1 = new Gecko(lizards1);
            gecko1.say();
        }
        else if(flag == 4){
            Reptiles reptiles2 = new Reptiles();
            Lizards lizards2 = new Lizards(reptiles2);
            MonitorLizard ml = new MonitorLizard(lizards2);
            ml.say();
        }
    }
}

解决方案

修改下你的Test类，提取成一个工厂类，面向抽象编程的原则，根据传入的type类型返回具体的某个实现类。
参考代码如下：

class AnimalFactory {
    public static Animal factory(int type) {
        Animal instance = null;
        switch (type) {
        case 1:
            Mammal mammal = new Mammal();
            instance = new Whale(mammal);
            break;
        case 2:
            Reptiles reptiles = new Reptiles();
            instance = new Lizards(reptiles);
            break;
        case 3:
            Reptiles reptiles1 = new Reptiles();
            Lizards lizards1 = new Lizards(reptiles1);
            instance = new Gecko(lizards1);
            break;
        case 4:
            Reptiles reptiles2 = new Reptiles();
            Lizards lizards2 = new Lizards(reptiles2);
            instance = new MonitorLizard(lizards2);
            break;
        default:
            System.out.println("not support this type.");
        }

        return instance;
    }

    public static void main(String[] args) {
        int flag = 4;
        Animal animal = AnimalFactory.factory(flag);
        if (animal != null) {
            animal.say();
        } else {
            System.out.println("this type is not supported.");
        }
    }

}

测试结果跟你的main方法一样，就是将你的main方法中new子类的过程作为工厂方法提出来，根据传入的类型来创建子类，最终返回的类型是抽象类。

解决方案二：

可以是可以，但是没必要。你的main提出出来，把 flag 作为参数，就是“简单工厂模式”。

http://baike.baidu.com/link?url=jB9egByWGlrT_MeCrx-KjOh_EV5SjeJ_8q6f-X-Nc77toPpSWSJ3OJ9wAO8JeyQkze75GDHri_Yhg4hnhLA8KK

解决方案三：

可以用抽象工厂模式！