Higher-kinded generics in Java

Question

Suppose I have the following class:

public class FixExpr {
  Expr<FixExpr> in;
}

Now I want to introduce a generic argument, abstracting over the use of Expr:

public class Fix<F> {
  F<Fix<F>> in;
}

But Eclipse doesn't like this:

The type F is not generic; it cannot be parametrized with arguments <Fix<F>>

Is this possible at all or have I overlooked something that causes this specific instance to break?

Some background information: in Haskell this is a common way to write generic functions; I'm trying to port this to Java. The type argument F in the example above has kind * -> * instead of the usual kind *. In Haskell it looks like this:

newtype Fix f = In { out :: f (Fix f) }

Answer 1

In order to pass a type parameter, the type definition has to declare that it accepts one (it has to be generic). Apparently, your F is not a generic type.

UPDATE: The line

F<Fix<F>> in;

declares a variable of type F which accepts a type parameter, the value of which is Fix, which itself accepts a type parameter, the value of which is F. F isn't even defined in your example. I think you may want

Fix<F> in;

That will give you a variable of type Fix (the type you did define in your example) to which you are passing a type parameter with value F. Since Fix is defined to accept a type parameter, this works.

UPDATE 2: Reread your title, and now I think you might be trying to do something similar to the approach presented in "Towards Equal Rights for Higher-Kinded Types" (PDF alert). If so, Java doesn't support that, but you might try Scala.

Answer 2

It looks as if you may want something like:

public class Fix<F extends Fix<F>> {
    private F in;
}

(See the Enum class, and questions about its generics.)

Answer 3

I think what you're trying to do is simply not supported by Java generics. The simpler case of

public class Foo<T> {
    public T<String> bar() { return null; }
}

also does not compile using javac.

Since Java does not know at compile-time what T is, it can't guarantee that T<String> is at all meaningful. For example if you created a Foo<BufferedImage>, bar would have the signature

public BufferedImage<String> bar()

which is nonsensical. Since there is no mechanism to force you to only instantiate Foos with generic Ts, it refuses to compile.

PS If I got this wrong, please correct me in the comments, don't just vote me down...

Answer 4

Maybe you can try Scala, which is a functional language running on JVM, that supports higher-kinded generics.

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[ EDIT by Rahul G ]

Here's how your particular example roughly translates to Scala:

trait Expr[+A]

trait FixExpr {
  val in: Expr[FixExpr]
}

trait Fix[F[_]] {
  val in: F[Fix[F]]
}