Re: [Camllist] Problem with polymorphic letrec
 TyngRuey Chuang
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Date:   (:) 
From:  TyngRuey Chuang <trc@i...> 
Subject:  Re: [Camllist] Problem with polymorphic letrec 
> ... Ignoring the base case of the recursion and the details of the > algorithm, I therefore need to code something like > > let rec polymul mul p q = > polymul (polymul mul) (fun _ > p) (fun _ > q) > > This expression does not typecheck in ocaml 2 because polymul is used > polymorphically within its own body. > > I could not come up with any hack to code polymul without changing the > representation of polynomials, which I'd rather not change for other > reasons. > > Any suggestion on how to work around this problem? > > Thanks in advance, > Matteo Frigo If you are certain the definition of a polymorphically recursive function f is typesafe, you can use "(Obj.magic f)" in the body of f to coerce the inferred type of f into one that is not constrained, so the definition of f can be typechecked. Note that Obj.magic has type Obj.magic: 'a > 'b so it in fact bypasses the type system. Use with care. :) Still I find it useful when defining map functions of "nested datatypes" like Nested.map below. Have fun! TyngRuey  module Pair = struct type 'a t = 'a * 'a let map f (x, y) = (f x, f y) end module Node = struct type ('a, 'b) t = Nil  Cons of 'a * 'b let map (f, g) x = match x with Nil > Nil  Cons (x, y) > Cons (f x, g y) end module Nested = struct type 'a t = Rec of ('a, 'a Pair.t t) Node.t let rec map f (Rec x) = Rec (Node.map (f, Obj.magic map (Pair.map f)) x) (*** HERE ***) end let n0 = Nested.Rec Node.Nil let n1 = Nested.Rec (Node.Cons (((1,2),(3,4)), n0)) let n2 = Nested.Rec (Node.Cons ((1,2), n1)) let n3 = Nested.Rec (Node.Cons (1, n2)) let double x = x + x let square x = x * x (* Some test cases *) let d3 = Nested.map double n3 let f3 = Nested.map string_of_int (Nested.map square n3)  Bug reports: http://caml.inria.fr/bin/camlbugs FAQ: http://caml.inria.fr/FAQ/ To unsubscribe, mail camllistrequest@inria.fr Archives: http://caml.inria.fr