Question

很难用语言解释我所追求的内容，但希望下面的代码示例带有注释就足够了。基本上我希望SubClass sc = new Subclass().method1()行返回Subclass实例。

public class SuperClass {

    public SuperClass method1()
    {
       //do whatever
       return this
    }
}

public class SubClass extends SuperClass {

    //we inherit method 1

    //method2
    public SubClass method2()
    {
       //do whatever
       return this
    }
}

//succesfully returns instance of Sublass, but...
SubClass sc = new Subclass().method2() 

//...the following line returns an instance of SuperClass and not Sublass
//I want Sublass's instance, without having to using overides
//Is this possible?

SubClass sc = new Subclass().method1()

编辑：----------------------------用例场景------------ -------------------

Message myMessage =  new ReverseTransactionMessageBuilder()
                    .policyNo(POLICY_NO) //on ReverseTransactionMessageBuilder
                    .audUserId(AUD_USER_ID) //on inherited MessageBuilder
                    .audDate(new Date()) //on inherited MessageBuilder
                    .processNo(EProcessConstants.FINANCE_MANUAL_ADJUSTMENT.getProcessCd()) //on inherited MessageBuilder
                    .serviceName("finance.ProcessReversalCmd") //on inherited MessageBuilder
                    .create(); //create is overridden so this is ReverseTransactionMessageBuilder

首先要注意的是sbrattla方式允许我以任何顺序调用这些.audDate（）。xxx（）方法。使用上面的类构造，你被迫最后调用sublcass上的方法（或使用一个非常丑陋的演员）

Answer 1

您需要执行以下操作：

public class SuperClass<T> {

  public T method1() {
    return (T) this;
  }

}

public class SubClass extends SuperClass<SubClass> {

  public SubClass method2() {
    return (SubClass) this;
  }

}

您可以在“Generics Introduction”中阅读有关Java Generics的更多信息，但简要说明您要告诉SuperClass将返回的实例强制转换为代表您定义的类型的T.在这种情况下，它是SubClass。

Answer 2

我认为您可以使用如下通用方法：

class Parent {
  public <T extends Parent> T instance() {
    return (T) this;
  }
}

class Child extends Parent {
}

class Test {
  public static void main() {
    Child child = new Parent().instance();
  }
}

父类返回子类

2 个答案: