I wrote:
> ... the Haskell report says that there is no way to declare an
> explicitly universally quantified type. My question is simple: why not?
I apologize: it is bad form to ask the same question twice, I know,
but I'm afraid my memory is failing -- I must be getting old ;-).
I forgot that we already had a discussion about this topic on this list
only a few months ago, and that discussion basically answered my question.
The answer, for anyone else who is interested, was that if you allow
nested quantifiers everywhere, then type-checking must solve
higher-order unification, which is in general undecidable. You can
allow quantifiers at the top level of data structures, though, and
according to Lennart Augustsson hbc allows that; so in hbc my example
could be rewritten as something like this:
data F = MkF t -> t -- did I get the syntax right?
foo :: (Int, String, F) -> (Int, String)
foo (i, s, MkF f) = (f i, f s)
--
Fergus Henderson <[EMAIL PROTECTED]> | "I have always known that the pursuit
WWW: <http://www.cs.mu.oz.au/~fjh> | of excellence is a lethal habit"
PGP: finger [EMAIL PROTECTED] | -- the last words of T. S. Garp.
