Java generics: parametricise a generic type

Question

Good Morning, I am working on some java code and i would like to make something like that:

public class MyClass <E, T extends MyInterface<T<E>>> { }

In this way, T should be both a generic and an instance of a class which implements MyInterface and should be parametric of the type E. But it seems it is not possible in Java, or maybe my approach to the problem is wrong. Thanks in advance.

Update:

First of all thanks to all who replied, now just to make things clear and let You understand what I am trying to achieve, I would like to make a generic container of generic TDAs. So let's say we need a container of Stacks based on Integers, then I would need to write something like that: MyContainer, of course i need to keep types separated for the insertion part, otherwise i could write MyContainer>. Moreover since all my TDAs implements MyInterface, I would like to set this as bound as well. So i guess

public class MyContainer < E, T extends MyInterface < E>>

should do the trick.

Possibly related to/duplicate of Simplifying Java class with 2 generic parameters. — Ben Schulz
– Ben Schulz, Commented Jun 13, 2012 at 10:59

Jon · Accepted Answer · 2012-06-13 11:04:22Z

1

This doesn't make sense.

public class MyClass <E, T extends MyInterface<T<E>>> { }

Because of the right-most T. Why would you want to parameterise a class with itself? I think what you really mean is this:

public MyClass<E, T extends MyInterface<E>>

Which essentially means that for every MyClass, you need an E, and a T MyInterface instance which is in turn parameterised to work with E.

I've found beginners can get very carried away with generics, which can result in code which is completely unreadable. Use generics sparingly, and think about whether you really need them in each given instance. If you are finding yourself writing code with multiple levels of nested generics parameterisation, then you're probably doing something wrong in terms of design.

answered Jun 13, 2012 at 11:04

Jon

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Comments

Piotr Praszmo · Accepted Answer · 2012-06-13 11:13:45Z

1

Basically you are trying to do something like this:

class MyClass<T<E>> // T has to be a generic class

or this:

class MyClass<T<Integer>> // T has to be a generic with Integer as parameter

You cannot and shouldnt have to do it, because it's meangingless. It doesn't tell you anything usefull about T. You could use any class as T like this:

DummyClass<T> extends Anything {};
MyClass<DummyClass<Integer>> myClass;

answered Jun 13, 2012 at 11:13

Piotr Praszmo

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JB Nizet · Accepted Answer · 2012-06-13 10:49:29Z

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I'm not sure if that's what you want, butI think it could just be

MyClass<E, T extends MyInterface<E>>

T<E> doesn't make much sense to me.

answered Jun 13, 2012 at 10:49

JB Nizet

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John B · Accepted Answer · 2012-06-13 11:33:35Z

0

Seems like it should be

public class MyClass <E, T extends MyInterface<E>> { }

Having public class MyClass <E, T extends MyInterface<T<E>>> { }

means that T

should be MyInterface<MyInterface<E>> which doesn't seem likely.

EDIT:

So if you need MyInterface<Class1<Class2>> it seems like it should be something like this:

public class MyClass <A, B extends SomeClass<A>, C extends MyInterface<B> { }

or

public class MyClass <E, T extends MyInterface<SomeOtherClass<E>>> { }

edited Jun 13, 2012 at 11:33

answered Jun 13, 2012 at 10:48

John B

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2 Comments

gr1ph00n Over a year ago

yes but in my idea, when a class which implements MyInterface should be instanced like so: MyInterface<Class1<Class2>>, that's why i wanted to add the double generic.

Jon Over a year ago

But you're saying that both "MyInterface" and Class1 need to implement MyInterface, and must be the same class. So if Foobar implements MyInterface, then you could do Foobar<Foobar<Class2>>. I can't think of any situations where that would be desirable.

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