Bulat Ziganshin wrote:
main = do return "xx" >>= ((\x -> print x) :: Show a => a -> IO ())
main2 = do return "xx" >>= (\(x:: (forall a . (Show a) => a)) -> print x)
main3 = do (x :: forall a . Show a => a) <- return "xx"
print x
the second and third variant should do the same, to my mind.
Well, you need to change your mind. :-) Forall is like a function; when you
write
forall a. Show a => a -> IO ()
it means something like
(a::Type) -> ShowDict a -> a -> IO ()
In other words, the caller supplies three things: the type a, a dictionary
for Show a, and a value of a. But when you write
forall a. Show a => a
it means something like
(a::Type) -> ShowDict a -> a
In other words, the caller supplies two things, the type a and a dictionary
for Show a, and the callee *returns* a value of a. In this case the callee
has to be prepared to produce a value of whatever type the caller requests,
and it can't do that if it only has a String.
On the other hand
exists a. Show a && a
means
(a::Type, ShowDict a, a)
i.e. the callee chooses the type and dictionary as well as the value, which
is what you need here.
-- Ben
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