[Haskell-cafe] A curios monad

[Ryan Ingram]

On Sat, Dec 13, 2008 at 3:00 PM, Andrew Coppin
<andrewcoppin at btinternet.com> wrote:
> So how is
>  foo :: ((forall b. a -> m b) -> m a) -> m a
> different from
>  bar :: forall b. ((a -> m b) -> m a) -> m a

Lets use a simpler example:
> foo :: (forall a. a -> a) -> (Int, String)
> bar :: forall a. (a -> a) -> (Int, String)

> -- this compiles
> foo f = (f 1, f "hello")

> -- this does not compile
> -- bar f = (f 1, f "hello")

> -- but this does
> bar f = (1, "hello")

The difference is that the *caller* of bar chooses how to instantiate
a, whereas the caller of foo must pass in a polymorphic function that
foo can instantiate at whatever type it wants... even multiple
different types!

> ident x = x
> plus1 x = x + 1 :: Int

> -- legal
> useFoo = foo ident
> -- not legal, not polymorphic
> -- useFoo = foo plus1

> -- legal
> useBar = bar ident
> -- also legal, instantiate "a" in the type of "bar" with "Int"
> useBar2 = bar plus1

  -- ryan