Java 有界类型

Question

我正在尝试理解 Java 中的有界类型。我认为我的推理几乎是正确的，但我得到了一个错误，我不明白为什么如果 A (JsonPerson) 是 T (Person) 的子类，它会给我这个错误。错误如下(代码中也有注释):

Error:(22, 16) error: incompatible types: JsonPerson cannot be converted to A
where A is a type-variable:
A extends Person declared in method <A>fromJson(JSONObject)

错误“发生”在返回行中。

我做了一个简单的例子，下面是代码

Person.java

public class Person {

    private String name;

    private String surname1;

    private String surname2;

    private String phone;

    private String email;

    public Person(String name, String surname1, String surname2, String phone, String email) {
        this.name = name;
        this.surname1 = surname1;
        this.surname2 = surname2;
        this.phone = phone;
        this.email = email;
    }

    public String getName() {
        return name;
    }

    public String getSurname1() {
        return surname1;
    }

    public String getSurname2() {
        return surname2;
    }

    public String getPhone() {
        return phone;
    }

    public String getEmail() {
        return email;
    }

}

JsonPerson.Java

public class JsonPerson extends Person implements JSONSerializableInterface<Person> {

    public JsonPerson(String name, String surname1, String surname2, String phone, String email) {
        super(name, surname1, surname2, phone, email);
    }

    /**
     * Error:(22, 16) error: incompatible types: JsonPerson cannot be converted to A
     * where A is a type-variable:
     * A extends Person declared in method <A>fromJson(JSONObject)
     * */
    @Override
    public <A extends Person> A fromJson(JSONObject json) throws JSONException {
        String name = json.getString("name");
        String surname1 = json.getString("surname1");
        String surname2 = json.getString("surname2");
        String phone = json.getString("phone");
        String email = json.getString("email");
        return new JsonPerson(name, surname1, surname2, phone, email);
    }

    @Override
    public JSONObject toJson(Person object) {
        return null;
    }
}

JSONSerializableInterdace.java

public interface JSONSerializableInterface<T> {

    public <A extends T> A fromJson(JSONObject json) throws JSONException;

    public JSONObject toJson(T object);
}

Answer 1

问题是有人可能会这样调用你的方法:

new JSONPerson().fromJSON<SpecialPerson>(json);

其中 SpecialPerson 扩展 Person。这是有效的(因为 A 扩展了 Person)，但是该方法返回的 JSONPerson 不是 SpecialPerson .