java - 如何基于某个类实例化具体的实现？

Question

我有几个扩展 bean 的对象:

public class Fruit {}

public class Banana extends Fruit {}

public class Pear extends Fruit {}

我有一个具有不同实现的接口(interface)，每个 bean 都有一个实现。

public interface Milkshake {
    public String doMilkshake();
}

public class FruitMilkshake implements Milkshake {
    public String doMilkshake() {
        return "Fruit Milkshake!";
    }
}

public class BananaMilkshake implements Milkshake {
    public String doMilkshake() {
        return "Banana Milkshake!";
    }
}

public class PearMilkshake implements Milkshake {
    public String doMilkshake() {
        return "Pear Milkshake!";
    }
}

如何根据 bean 的具体类型实例化正确的实现？

现在我已经使用了类型化和 Map 来“映射”正确的实现。像这样:

public void hungry(Fruit fruit) {
    Map<String, String> obj2impl = new HashMap<String, String>();
    obj2impl.put("Fruit", "FruitMilkshake");
    obj2impl.put("Banana", "BananaMilkshake");
    obj2impl.put("Pear", "PearMilkshake");

    String name = fruit.getClass().getCanonicalName();
    String implName = obj2impl.get(name);
    Milkshake milkshake = (Milkshake) Class.forName(implName).newInstance();

    milkshake.doMilkshake(fruit);
}



public interface Milkshake <T t> {
    public String doMilkshake(T t);
}

public class FruitMilkshake implements Milkshake<Fruit> {
    public String doMilkshake(Fruit fruit) {
        return "Fruit Milkshake!";
    }
}

public class BananaMilkshake implements Milkshake<Banana> {
    public String doMilkshake(Banana banana) {
        return "Banana Milkshake!";
    }
}

public class PearMilkshake implements Milkshake<Pear> {
    public String doMilkshake(Pear pear) {
        return "Pear Milkshake!";
    }
}

实现这一目标的更好方法？

Answer 1

Better ways to achieve this?

首先是避免反射(reflect)类的名称:

Map<String, Class<? extends Milkshake>> obj2impl =
    new HashMap<String, Class<? extends Milkshake>>();
obj2impl.put("Fruit", FruitMilkshake.class);
obj2impl.put("Banana", BananaMilkshake.class);

...

Milkshake milkshake = obj2impl.get(text).newInstance();

现在仍然要求您在每个实现中都有一个无参数构造函数，并且它仍然总是创建一个新实例。如果您使用 Provider -类似的概念，你可以绕过这个:

Map<String, Provider<Milkshake>> map = ...;
// Fill the map with providers, some of which could create a new instance,
// and some could reuse an existing one

...

Milkshake milkshake = map.get(text).get();

编辑:重读这篇文章后，您也可以删除文本部分，并得到 Map<Class<?>, Provider<Milkshake>> 。如果可能的话，避免对类的名称进行硬编码。

当然，如果你的Fruit类(class)有一个makeMilkshake抽象方法，那就更好了...