Overriding type parameters

Question

I'd like to implement the following classes with the following hierarchy:

public class DwellingPropertySetter
    extends AbstractPropertySetter<Dwelling>

public class HousePropertySetter
    extends DwellingPropertySetter<House>

public class SkyscrapperPropertySetter
    extends HousePropertySetter<Skyscrapper>

Unfortunately this code won't compile. A way to do it would be this:

public class DwellingPropertySetter<T extends Dwelling>
    extends AbstractPropertySetter<T>

public class HousePropertySetter<T extends House>
    extends DwellingPropertySetter<T>

public class SkyscrapperPropertySetter<T extends Skyscrapper>
    extends HousePropertySetter<T>

But for me those extends keywords are unnecessary. Whenever I want to use a SkyscrapperPropertySetter I'd have to specify a type parameter. This is useless and would then look like SkyscrapperPropertySetter<Skyscraper>.

Do you know a way out for me? The only other way I know to realise my first hierarchy would be using interfaces and delegate methods.

Why do you want to use generics (like in your second example) when your first example seems to do what you need? — assylias, May 11 '12 at 12:03
@NewTalk, here's a potentially stupid question: Why doesn't it compile? — aioobe, May 11 '12 at 12:07
@aioobe Because of _The type DwellingPropertySetter is not generic; it cannot be parameterized with arguments _ — , May 11 '12 at 12:08

score 2 · Accepted Answer · answered May 11 '12 at 12:16

I think you have correctly recognized it is pointless to have something like ChickenArrayList<Chicken>, so you can either create something like

class ChickenArrayList extends ArrayList<Chicken>

OR, if you want to reuse some functionality in ChickenArrayList, you may have to make it abstract (and generic) and put another concrete, non-generic class on top of it:

class AbsChickenArrayList<T extends Chicken> extends ArrayList<T>
// and
class ChickenArrayList extends AbsChickenArrayList<Chicken>
class HenArrayList extends AbsChickenArrayList<Hen>

I know this is quite verbose, but this is the best you can do with this ~18 year old language.

Marko Topolnik · Answer 2 · 2012-05-11T12:15:36.247

1

I think you can achieve what is reasonable. If you have

public class DwellingPropertySetter
extends AbstractPropertySetter<Dwelling>

That means you have already made DwellingPropertySetter non-generic, all the method signatures will use Dwelling. If you say

public class DwellingPropertySetter<T extends Dwelling>
extends AbstractPropertySetter<T>

that means there can be various DwellingPropertySetters -- and you really would like to have different ones, so I think this is what you really want to keep. Or is it your complaint that you want both a DwellingPropertySetter with no args, where Dwelling is assumed, AND a subclass extending DwellingPropertySetter<T> with a different T? That cannot be had with Java Generics.

edited May 11 '12 at 12:15

answered May 11 '12 at 11:56

Marko Topolnik

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Yes. I want to have a `DwellingPropertySetter` with no arguments and and a subclass extending `DwellingPropertySetter` with a different `T`. Why isn't that technically possible? :-( – May 11 '12 at 12:11
In Java Generics, but also in every other implementation of the same concept in other languages, a type either does or doesn't take type parameters. I've never seen a case where a type param would be optional with a default value. EDIT I googled a bit, C++ actually [does support](http://stackoverflow.com/questions/707780/is-there-a-reasonable-approach-to-default-type-parameters-in-c-sharp-generics) such a feature. – Marko Topolnik May 11 '12 at 12:13
@NewTalk Nothing prevents you from writing `SkyscrapperPropertySetter sps = new SkyscrapperPropertySetter()` without the generics. – assylias May 11 '12 at 12:16
2

@assylias Only if he disables compiler warnings, otherwise the screen is going to crawl with yellow squiggles :) – Marko Topolnik May 11 '12 at 12:19

Overriding type parameters

2 Answers2