是否可以将引用类型传递给需要动态引用类型的对象实例？

3回答

慕尼黑5688855

假设您有以下POJO模型：class Car {}class Lamborghini extends Car {}class Ferrari extends Car {}您的Manager课程如下所示：class Manager<T extends Car> {    private Class<T> clazz;    public Manager(Class<T> clazz) {        this.clazz = clazz;    }    public String generateCustomCarName() {        if (Lamborghini.class == clazz) {            return "Lambooooo!";        } else if (Ferrari.class == clazz) {            return "Wild horse Ferrari!";        }        return "Not for everyone!";    }    @Override    public String toString() {        return String.format("Manager[clazz=%s]", clazz.getName());    }}现在，让我们创建一个Store有 oneFerrari和 one的Lamborghini：class Store {    private Car ferrari = new Ferrari();    private Car lamborghini = new Lamborghini();    public Car getFerrari() {        return ferrari;    }    public void setFerrari(Car ferrari) {        this.ferrari = ferrari;    }    public Car getLamborghini() {        return lamborghini;    }    public void setLamborghini(Car lamborghini) {        this.lamborghini = lamborghini;    }}现在，我们需要class在运行时获取信息。由于Java's Type Erasure我们需要提供class：class CustomCarSerializer<T extends Car> extends StdSerializer<T> {    public CustomCarSerializer(Class<T> clazz) {        super(clazz);    }    @Override    public void serialize(T value, JsonGenerator jgen, SerializerProvider provider)        throws IOException {        Manager<T> manager = new Manager<>(_handledType);        jgen.writeString(manager.generateCustomCarName());    }}请注意，它StdSerializer也有需要的构造函数，class我们可以稍后使用_handledType. 现在我们需要使用序列化器配置我们的实现SimpleModule：import com.fasterxml.jackson.core.JsonGenerator;import com.fasterxml.jackson.databind.ObjectMapper;import com.fasterxml.jackson.databind.SerializationFeature;import com.fasterxml.jackson.databind.SerializerProvider;import com.fasterxml.jackson.databind.module.SimpleModule;import com.fasterxml.jackson.databind.ser.std.StdSerializer;import java.io.IOException;public class Cars {    public static void main(String[] args) throws Exception {        SimpleModule carsModule = new SimpleModule("CarsModule");        carsModule.addSerializer(Lamborghini.class, new CustomCarSerializer<>(Lamborghini.class));        carsModule.addSerializer(Ferrari.class, new CustomCarSerializer<>(Ferrari.class));        ObjectMapper mapper = new ObjectMapper();        mapper.registerModule(carsModule);        mapper.enable(SerializationFeature.INDENT_OUTPUT);        System.out.println(mapper.writeValueAsString(new Store()));    }}上面的代码打印：{  "ferrari" : "Wild horse Ferrari!",  "lamborghini" : "Lambooooo!"}

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人到中年有点甜

处理这个问题的正确方法是让两个类（例如Car和Desk）实现一个公共接口（例如Item）并将该接口与Manager该类相关联。public class Instance {    private static interface Item {        String getName();    }    private static class Car implements Item {        @Override        public String getName() {            return "Car";        }    }    private static class Desk implements Item {        @Override        public String getName() {            return "Desk";        }    }    private static class Manager<T extends Item> {        private T item;        public Manager(T item) {            this.item = item;        }        @Override        public String toString() {            return String.format("Manager[item=%s]", item.getName());        }    }    public static <E extends Item> Manager<E> buildItemManager(E item) {        return new Manager<E>(item);     }    public static void main(String[] args) {        Item car  = new Car(), desk = new Desk();        Manager<Item> manager1 = buildItemManager(car);        System.out.println(manager1);        Manager<Item> manager2 = buildItemManager(desk);        System.out.println(manager2);    }}输出Manager[item=Car]Manager[item=Desk]另一种方法...Manager通过传入一个代表 a 的类来实例化 a Item。public class Instance {    public static interface Item {        String getName();    }    public static class Car implements Item {        @Override        public String getName() {            return "Car";        }    }    public static class Desk implements Item {        @Override        public String getName() {            return "Desk";        }    }    public static class Manager<T extends Item> {        private T item;        public Manager(T item) {            this.item = item;        }        @Override        public String toString() {            return String.format("Manager[item=%s]", item.getName());        }    }    public static <E extends Item> Manager<E> buildItemManager(Class<E> item) throws InstantiationException, IllegalAccessException {        return new Manager<E>(item.newInstance());     }    public static void main(String[] args) {        try {            /* Cannot initialize array with generic types... */            @SuppressWarnings("unchecked")            Manager<? extends Item>[] managers = new Manager[2];            managers[0] = buildItemManager(Car.class);            managers[1] = buildItemManager(Desk.class);            for (Manager<?> manager : managers) {                System.out.println(manager);            }        } catch (InstantiationException e) {            e.printStackTrace();        } catch (IllegalAccessException e) {            e.printStackTrace();        }    }}

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qq_笑_17

是的，您可以使用泛型。  public <T> void builCarManager(T car) {    Manager<T> carManager = new Manager<>(); }现在您可以传递任何类型的对象。创建的管理器将属于同一类型。

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