Implementation of generic interface with specific type in Java

Question

Consider the following generic interface:

interface Petlover<T>{
    void train(T somePet);
}

I understand that it's possible to provide generic implementations to generic interfaces (e.g., class MalePetLover<T> implements Petlover<T>). But I am having trouble implementing the interface for a specific type only:

class Dogperson implements Petlover<T> {
    int age;

    void train(String someDog){...};
}

The compilation error is train(String) in Dogperson cannot implement train(T) in Petlover. What would be a right way to handle this?

implements PetLover<String>? You should clarify what you are trying to achieve here — user180100
– user180100, Commented May 14, 2017 at 13:05
The String is intended to be the name of a dog. I tried Petlover <String> to no avail (please see Eran's answer below) - the same error is output by the compiler. — flow2k
– flow2k, Commented May 14, 2017 at 13:13

Eran · Accepted Answer · 2017-05-14 13:13:30Z

6

Since you expect train to accept a String, your class should implement Petlover<String>:

class Dogperson implements Petlover<String> {
    int age;

    public void train(String someDog) {...};
}

or perhaps the train() method of Dogperson should accept a Dog argument, and then the class would implement Petlover<Dog>.

edited May 14, 2017 at 13:13

answered May 14, 2017 at 13:06

Eran

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6 Comments

flow2k Over a year ago

I seem to be getting the same error after changing T to String as you suggest: error: train(String) in Dogperson cannot implement train(T) in Petlover. Yes, it would be better to use a Dog object, but I just kept it as String for simplicity.

Eran Over a year ago

@flow2k It passed compilation for me. You just have to add public access modifier to the train() method.

flow2k Over a year ago

Ah - okay. I missed that in your answer. Thanks.

Lew Bloch Over a year ago

Side note: You should follow the Java naming conventions.

flow2k Over a year ago

@LewBloch Do you mean CamelCase?

|

davidxxx · Accepted Answer · 2017-05-14 13:22:16Z

3

The subclass that extends a generics class has to conform to the type that it specifies in its own declaration.

You declare : class Dogperson implements Petlover<T> {, so the train() method has to have this signature : void train(T someDog){...};

If you want to have this signature in the subclass:

void train(String someDog){...};

you have to declare a subclass that implements PetLover by specifying the String class as parameterized type :

class Dogperson implements Petlover<String> {
     public void train(String someDog){...};
}

You can specify any parameter as T derives from Object in the interface.
For example, if you want to have a Dog as type used in the subclass, you could write it :

class Dogperson implements Petlover<Dog> {
     public void train(Dog dog){...};
}

For example, if you want to have a Cat, you could write it :

class Catperson implements Petlover<Cat> {
     public void train(Cat cat){...};
}

edited May 14, 2017 at 13:22

answered May 14, 2017 at 13:07

davidxxx

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2 Comments

Erwin Bolwidt Over a year ago

Like Eran mentioned in the other answer, the train methods in the Dogperson and Catperson classes need to be public.

davidxxx Over a year ago

@Erwin Bolwidt True. Interface methods has to be public. We would have seen it before.

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