Higher-kinded generics in Java
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) }
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 Foo
s with generic T
s, it refuses to compile.
Maybe you can try Scala, which is a functional language running on JVM, that supports higher-kinded generics.
[ 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]]
}
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.
Still, there are ways to encode higer-kinded generics in Java. Please, have a look at higher-kinded-java project.
Using this as a library, you can modify your code like this:
public class Fix<F extends Type.Constructor> {
Type.App<F, Fix<F>> in;
}
You should probably add an @GenerateTypeConstructor
annotation to your Expr
class
@GenerateTypeConstructor
public class Expr<S> {
// ...
}
This annotation generates ExprTypeConstructor class. Now you can process your Fix of Expr like this:
class Main {
void run() {
runWithTyConstr(ExprTypeConstructor.get);
}
<E extends Type.Constructor> void runWithTyConstr(ExprTypeConstructor.Is<E> tyConstrKnowledge) {
Expr<Fix<E>> one = Expr.lit(1);
Expr<Fix<E>> two = Expr.lit(2);
// convertToTypeApp method is generated by annotation processor
Type.App<E, Fix<E>> oneAsTyApp = tyConstrKnowledge.convertToTypeApp(one);
Type.App<E, Fix<E>> twoAsTyApp = tyConstrKnowledge.convertToTypeApp(two);
Fix<E> oneFix = new Fix<>(oneAsTyApp);
Fix<E> twoFix = new Fix<>(twoAsTyApp);
Expr<Fix<E>> addition = Expr.add(oneFix, twoFix);
process(addition, tyConstrKnowledge);
}
<E extends Type.Constructor> void process(
Fix<E> fixedPoint,
ExprTypeConstructor.Is<E> tyConstrKnowledge) {
Type.App<E, Fix<E>> inTyApp = fixedPoint.getIn();
// convertToExpr method is generated by annotation processor
Expr<Fix<E>> in = tyConstrKnowledge.convertToExpr(inTyApp);
for (Fix<E> subExpr: in.getSubExpressions()) {
process(subExpr, tyConstrKnowledge);
}
}
}
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.)