PECS:如何将消费者转变为生产者？

Question

我有一家供应餐食的餐馆。厨房得到的盘子是消费者的。

class Food{}
class Bamboo extends Food {}

interface Kitchen {
    void build(List<? super Food> dessert);
}

abstract class Restaurant {
    Kitchen kitchen;

    public Restaurant(Kitchen kitchen) {
        this.kitchen = kitchen;
    }

    List<? extends Food> getMeals() {
        List<Food> food = new ArrayList<>();
        this.kitchen.build(food);
        return food;
    }
}

class PandaKitchen implements Kitchen{

    // Exact signature of the Kitchen
    @Override
    public void build(List<? super Food> plate) {
        // the List IS a consumer of bamboos
        plate.add(new Bamboo());
    }
}

// Bamboo specialized restaurant
class OhPanda extends Restaurant {

    public OhPanda() {
        super(new PandaKitchen());
    }

    // Specialized signature of List<? extends Food>
    @Override
    List<Bamboo> getMeals() {
        List<? super Food> bamboos = new ArrayList<>();
        this.kitchen.build(bamboos);

        // Obviously here, there is no information of having only Bamboos
        return bamboos; // <==== FAIL

        //return (List<Bamboo>) bamboos; // would not compile
    }
}

在最后一行中，我知道我的OhPanda餐厅只生产竹子。在不在内存中创建/复制ArrayList的情况下转换List<? super Food>的最佳实践是什么？

更完整的要点写在这里：https://gist.github.com/nicolas-zozol/8c66352cbbad0ab67474a776cf007427

Answer 1

或者，也许你可以写一个餐厅和厨房的打字版本？

package kitchen;

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

class Food{}
class Bamboo extends Food {}

interface Kitchen<F> {
    void build(List<F> dessert);
}

abstract class Restaurant<T> {
    protected Kitchen kitchen;

    Restaurant(Kitchen kitchen) {
        this.kitchen = kitchen;
    }

    List<T> getMeals() {
        List<T> food = new ArrayList<>();
        kitchen.build(food);
        return food;
    }
}

class PandaKitchen implements Kitchen<Bamboo>{

    @Override
    public void build(List<Bamboo> dessert)
    {
        dessert.add(new Bamboo());
    }
}

/** Bamboo specialized restaurant*/
class OhPanda extends Restaurant<Bamboo> {

    OhPanda() {
        super(new PandaKitchen());
    }

    @Override
    List<Bamboo> getMeals() {
        List<Bamboo> bamboos = new ArrayList<>();
        kitchen.build(bamboos);
        return bamboos;
    }
}

Answer 2

我认为您错误地使用了下限通配符。通过在实现中不指定通配符的类，您可以使用它来转向您可能面临的上限通配符限制。我不认为你想和食物及其超级类型打交道。你只想使用食品和它的衍生产品，你应该找到一个“扩展食品”的解决方案，甚至去掉通配符，使用List<食品>