Why does the type parameter bound T <: Comparable[T] fail for T = Int?

Question

scala> class Foo[T <: Comparable[T]](val x : T)
defined class Foo

scala> (3: Int).asInstanceOf[Comparable[Int]]  
res60: java.lang.Comparable[Int] = 3

scala> new Foo(3)                              
<console>:13: error: inferred type arguments [Int] do not conform to class Foo's type parameter bounds [T <: java.lang.Comparable[T]]
       new Foo(3)
       ^

Is the 2nd expression the result of type erasure?

How would I go about defining Foo so that I could parameterize it with Int but still be able to perform some ordering behavior with its instance variable?

Answer 1

Use a view bound.

Welcome to Scala version 2.8.0.final (Java HotSpot(TM) Client VM, Java 1.6.0_21).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class Foo[T <% Comparable[T]](val x : T)
defined class Foo

scala> new Foo(3)
res0: Foo[Int] = Foo@9aca82

Answer 2

The question, as stated, is still unanswered (though "use view bounds" solves the problem, which is more useful). The answer is simply that an Int in Scala is supposed to be equivalent to an int in Java, which is not a class at all, and, therefore, cannot even be a Comparable (though that could be solved in Java 7 with defender methods... I wonder if they'll do it).

The solution given, to use a view bound, is used throughout Scala to solve the problem of a class that could implement something but doesn't, because it is not under Scala's control -- ie, Java classes.

And, of course, it can be used by programmers themselves to deal with similar stuff from libraries and frameworks, or simply to produce wrappers around a library to give it a Scala-ish feeling.