[Haskell-cafe] Rank N type tutorial?

[rossberg at ps.uni-sb.de]

"Greg Buchholz" <haskell at sleepingsquirrel.org> wrote:
> I'm not quite sure why this is illegal...
>
>> foo :: Integer -> (forall a. Show a => a)
>> foo 2 = ["foo"]
>> foo x = x

The type signature promises that foo returns a function that has type a
*for all a* (that are in Show). But neither a string list nor an integer
can have all types. Rather, they are very specific. That is, what you have
implemented actually could have type

foo :: Integer -> (exists a. Show a => a)

(if GHC supported such types).

Note that only bottom can have a type like forall a.a.

> ...while this is just fine...
>
>> bar :: Integer -> (forall a. Show a => a->b) -> b
>> bar 2 k = k ["bar"]
>> bar x k = k x

Here, you declare bar to *take* a function that delivers some b for any a.
That is, the burden of defining such a function now is on the caller, not
bar itself.

However, in this case, defining such an argument function actually is
easy, for another reason: the signature says k returns a result of type b,
but b is fixed. So it is always possible to come up with a trivial
instance of such a function (which will just ignore its argument). For
instance,

bar 2 (\_ -> "boo")

But try to call this for a different example:  ;-)

baz :: Integer -> (forall a b. Show a => a->b) -> c
baz 2 k = k ["baz"]
baz x k = k x

Hope this helps,
- Andreas