GHC.Prim.Any's created instead of free variable [in polymorphic function]

Question

Is this a bug?

{-# LANGUAGE NoMonomorphismRestriction #-}
import qualified Text.Parsec.Token as P
import Text.Parsec.Language (haskellDef)
(P.TokenParser { P.identifier = ident }) = P.makeTokenParser haskellDef

yields ident of type Text.Parsec.Prim.ParsecT String GHC.Prim.Any Data.Functor.Identity.Identity String, whereas defining

haskell = P.makeTokenParser haskellDef
ident = P.identifier haskell

yields one of type Text.Parsec.Prim.ParsecT String u Data.Functor.Identity.Identity String

edit

The behavior is not identical in ghci,

infixl 4 <++>
(<++>) = liftM2 (++)

(P.TokenParser { P.identifier = ident }) = P.makeTokenParser haskellDef
suitable_macro = ident

parseMacro = many space *> suitable_macro

parseMacro' =
    try (string "{{" *> parseMacro <* string "}}")

parseAll = many (noneOf "{") <++>
    option "" (parseMacro' <|> (string "{" <++> parseAll))

Then, try to run it,

*Hz2.Preproc> parseTest parseAll "asdf{{b}}"

<interactive>:0:11:
    Couldn't match expected type `()' with actual type `GHC.Prim.Any'
    Expected type: Parsec String () a0
    Actual type: ParsecT
                    String GHC.Prim.Any Data.Functor.Identity.Identity [Char]
    In the first argument of `parseTest', namely `parseAll'
    In the expression: parseTest parseAll "asdf{{b}}"

Answer 1

Not really; I believe it's Report-compliant behaviour: fully polymorphic type variables in patterns are instantiated to Any. However, in GHC 7.2 onwards, this works like you'd expect (see especially the commit message at the end).

As for the GHCi behaviour, this is because GHCi's extended defaulting rules default fully-polymorphic variables to ().