[Haskell-cafe] A curios monad
dave at zednenem.com
Sun Dec 14 02:50:28 EST 2008
On Sat, Dec 13, 2008 at 6:00 PM, Andrew Coppin
<andrewcoppin at btinternet.com> wrote:
> David Menendez wrote:
>> On Thu, Dec 11, 2008 at 1:48 PM, Andrew Coppin
>> <andrewcoppin at btinternet.com> wrote:
>>> BTW, does anybody know how rank-N types are different from existential
>> You mean the Haskell extensions?
>> ExistentialQuantification lets you define types such as,
>> data SomeNum = forall a. Num a => SomeNum a
>> RankNTypes lets you nest foralls arbitrarily deep in type signatures,
>> callCC :: ((forall b. a -> m b) -> m a) -> m a -- this is rank-3
>> RankNTypes implies ExistentialQuantification (among others).
> 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
Daniel Fischer already gave the short answer, I'll try explaining why
someone might *want* a rank-3 signature for callCC. It involves a
continuation monad, but hopefully nothing headache-inducing.
The type for callCC in Control.Monad.Cont.Class is,
callCC :: forall m a b. (MonadCont m) => ((a -> m b) -> m b) -> m b
You can use callCC to do very complicated things, but you can also use
it as a simple short-cut escape, like the way "return" works in C.
foo = callCC (\exit -> do
x <- if something
then return 'a'
else exit False
The type of exit is Bool -> m Char, which is fine in this example, but
it means that we can only use exit to escape from computations that
are producing characters. For example, we cannot write,
bar = callCC (\exit -> do
x <- if something then return 'a' else exit False
y <- if something then return 42 else exit False
Because exit would need to have the type Bool -> m Char the first time
and Bool -> m Int the second time. But, if callCC had a rank-3 type,
callCC :: forall m a. (MonadCont m) => ((forall b. a -> m b) -> m a) -> m a
then exit would have the type "forall b. Bool -> m b", and bar would
compile just fine.
Dave Menendez <dave at zednenem.com>
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