How can I combine the typeclass pattern with subty

问题:

Suppose I'm using the typeclass pattern in Scala. Here's how I make a class C part of the typeclass Foo:

Welcome to Scala version 2.9.0.1 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_26).

scala> trait Foo[T] { def foo(t: T) }
defined trait Foo

scala> def foo[T : Foo](t: T) { implicitly[Foo[T]].foo(t) }
foo: [T](t: T)(implicit evidence$1: Foo[T])Unit

scala> class C
defined class C

scala> foo(new C)
<console>:11: error: could not find implicit value for evidence parameter of type Foo[C]
       foo(new C)
          ^

scala> implicit object FooC extends Foo[C] { override def foo(c: C) { println("it's a C!") } }
defined module FooC

scala> foo(new C)
it's a C!

So far so good. But suppose I have a subclass D of C, and I want instances of D to be "in" the typeclass too:

scala> class D extends C
defined class D

scala> foo(new D)
<console>:13: error: could not find implicit value for evidence parameter of type Foo[D]
       foo(new D)
          ^

Doh! How do I make this work without having to explicitly provide a typeclass instance for D?

回答1:

There are different possible solutions for this, depending on whether I want to fix the problem only for C, or whether I want to fix the problem for the entire typeclass.

For C only, instead of implicit object FooC ... we say:

implicit def CIsFoo[T <: C]: Foo[T] =
  new Foo[T] { override def foo(t: T) { println("it's a C!") } }

To fix all of Foo, make it contravariant:

trait Foo[-T] { def foo(t: T) }

Or if for some reason you can't or don't want to do that, you can replace def foo... with:

def foo[T](t: T)(implicit foo: Foo[_ >: T]) =
  foo.foo(t)

(Thanks to #scala denizens Daniel Sobral and Stefan Zeiger for their help.)

UPDATED Sep 20 2011 to include the "make Foo contravariant" solution, which I missed