Robert Greayer wrote:
f0 _ = (foo True, foo 'x') where foo = id
is well-typed.
Really? That actually works? How interesting... This suggests to me that
where-clauses also do strange things to the type system.
whereas
f1 foo = (foo True, foo 'x')
requires 'foo' to be polymorphic in its first argument. This does
require a higher rank type, which can't be inferred:
You could type f1 as
f1 :: (forall a . a -> a) -> (Bool, Char)
and apply it to 'id'.
Or you could type it as something like:
f1 :: (forall a . a -> ()) -> ((),())
and apply it to 'const ()'
...all of which is beyond Haskell-98, which is what I am limiting myself
to at present.
(Actually, even that is a lie. I don't have type-classes yet...)
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