I'm obviously missing something here. I dont understand what monomorphism has to do with the given example as the implicit parameter would be the same type [a] for some type a in each case. If we made the parameter explicit then removing the type definition would not cause this problem. Are implicit parameters not simply made explict by the compiler? -----Original Message----- From: John Hughes [mailto:rjmh@cs.chalmers.se] Sent: 04 February 2002 10:26 To: haskell@haskell.org; koen@cs.chalmers.se Subject: Re: Implicit Parameters Suppose I have the following function definition: app :: (?ys :: [a]) => [a] -> [a] app xs = case ?ys of [] -> xs (y:ys') -> y : (app xs with ?ys = ys') This function appends its argument to its implicit argument, by recursively changing the implicit argument. For example: Hugs> app [1,2] with ?ys = [3,4] [3,4,1,2] So far so good! Now, since Haskell has type inference, we can leave out the type signature: -- app :: (?ys :: [a]) => [a] -> [a] app xs = case ?ys of [] -> xs (y:ys') -> y : (app xs with ?ys = ys') Let us check if it still works again: Hugs> app [1,2] with ?ys = [3,4] [3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,{Interrupted!} And, stunningly, it doesn't! Why doesn't this work? That is because type inference assumes that the body of `app' is monomorphic, i.e. has no context in its type. Therefore, the recursive call to app where ?ys is changed, had no effect at all. It works with the type signature there, because in order to implement type checking (note: not type inference) with polymorphic recursion demands to use the stated type in checking the body of the function. Mark Shields, Simon Peyton-Jones, and I, and also Jörgen Gustavsson have been discussing several modest solutions to this (we believe it is not needed to do full type inference that can deal with polymorphic recursion to solve this problem). Not so fast! Proposing a "solution" means this is regarded as a "problem"! But what is to say that the first behaviour is "right" in any general sense? The important thing is that the language semantics is clear, and this is a semantic question. The static semantics of Haskell *is* clear: recursive calls are monomorphic unless a type signature is given; this is the basis for understanding the behaviour above. When implicit parameters are used, it's very important to be aware whether a binding is monomorphic or not (can't resist plugging := again!). Will your "solution" make understanding when a binding is monomorphic simpler? If not, it could be worse than the "problem" -- and the fact that it makes this one example behave as you want is no justification. John _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell