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Home/ Questions/Q 6904807
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T08:06:24+00:00 2026-05-27T08:06:24+00:00

I am trying to create a generic Identifier class which I would be able

  • 0

I am trying to create a generic Identifier class which I would be able to use as follows:

public class TestGenericIdentifier {
    public static void main(String[] args) {
        Identifier<Car> carId = new Identifier<>(Car.IdentifierType.LICENSE_PLATE, "123 XYZ");
        Identifier<Person> personId = new Identifier<>(Person.IdentifierType.SOCIAL_SECURITY, "123456");
        System.out.println(carId);
        System.out.println(personId);
    }
}

To get there, I started by creating an Identifiable interface:

public interface Identifiable<T extends Enum> {}

The idea being that a class that implements Identifiable needs to provide an enum T in its declaration which is the type of the first parameter of the Identifier constructor:

public class Identifier<E extends Identifiable<T>> { //does not compile
    public Identifier(T type, String value) {
        //some code
    }
}

Now the code above does not compile as I can only use Identifiable (no parameter T) on the first line. If it worked I would be able to write the following two classes:

public class Car implements Identifiable<Car.IdentifierType>{

    public enum IdentifierType {
        SERIAL_NUMBER,
        LICENSE_PLATE;
    }
}

public class Person implements Identifiable<Person.IdentifierType> {

    public enum IdentifierType {
        DATABASE_ID,
        SOCIAL_SECURITY;
    }
}

Is there a way to do that using generics?

EDIT
One way is to compromise conciseness and keep compile-time type checking by doing:

public class Identifier<T extends Enum> {

    public Identifier(T type, String value) {
    }
}

and the main function becomes:

Identifier<Car.IdentifierType> carId = new Identifier<>(Car.IdentifierType.LICENSE_PLATE, "123 XYZ");
Identifier<Person.IdentifierType> personId = new Identifier<>(Person.IdentifierType.SOCIAL_SECURITY, "123456");
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T08:06:25+00:00Added an answer on May 27, 2026 at 8:06 am

    You can get this to compile by tweaking your code a bit but I’m not sure it’s what you want. The following seems to work for me.

    Identifier<Car.IdentifierType, Car> carId =
        new Identifier<Car.IdentifierType, Car>(Car.IdentifierType.LICENSE_PLATE,
            "123 XYZ");
    
    public static class Identifier<T extends Enum, E extends Identifiable<T>> {
        public Identifier(T type, String value) {
            // some code
        }
    }
    

    The question is why do you want to do this? If you edit your question some more with the background, I can edit my answer to be more helpful.

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