User:Benmachine/Cont

I found a bug with the <hask> tag. I put it on its own page so it doesn't ruin my user page.

A practical Cont tutorial

It seems to me like Cont and ContT is way simpler than people make it. Essentially what it seems to be is a way to give a name to the "tail" of a do-block.

contstuff :: Magic a
contstuff = do
  thing1
  thing2
  -- Here I want to manipulate the rest of the computation.
  -- So I want a magic function that will give me the future to
  -- play with.
  magic $ \rest ->
    -- 'rest' is the rest of the computation. Now I can just do it,
    -- or do it twice, or discard it, or
    -- do it and then use the result to do it again... it's easy to
    -- imagine why this might be useful.
  thing3 -- these might get done once, several times,
  thing4 -- or not at all.

The question is, what type should magic have? Well, let's say the whole do-block results in a thing of type r (without thinking too hard about what this means). Then certainly the function we give magic should result in type r as well, since it can run that do-block. The function should also accept a single parameter, referring to the tail of the computation. That's the rest of the do-block, which has type r, right? Well, more or less, with one caveat: we might bind the result of magic:

  x <- magic $ \rest -> -- ...
  thingInvolving x

so the rest of the do-block has an x in it that we need to supply (as well as other variables, but magic already has access to those). So the rest of the do-block can be thought of as a bit like a -> r. Given access to the rest of that do-block, we need to produce something of type r. So our lambda has type (a -> r) -> r and hence magic :: (a -> r) -> r -> Magic a

Magic a = Cont r a
magic = Cont

Tada!

The thing with Cont is I could implement it way before I understood it, because the types have really only one implementation, but here's a way of using the intuition above to implement Functor without thinking about the types too much:

instance Functor (Cont r) where
  fmap f (Cont g) = -- ...

Well, we've got to build a Cont value, and those always start the same way:

  fmap f (Cont g) = Cont $ \rest -> -- ...

Now what? Well, remember what g is. It looks like \rest -> stuffWith (rest val), where val is the 'value' of the computation (what would be bound with <-). So we want to give it a rest, but we don't want it to be called with the 'value' of the computation - we want f to be applied to it first. Well, that's easy:

  fmap f (Cont x) = Cont $ \rest -> x (\val -> rest (f val))

Load it in `ghci` and the types check. Amazing! Emboldened, let's try Applicative

instance Applicative (Cont r) where
  pure x = Cont $ \rest -> -- ...

We don't want to do anything special here. The rest of the computation wants a value, let's just give it one:

  pure x = Cont $ \rest -> rest x

What about <*>?

  Cont f <*> Cont x = Cont $ \rest -> -- ...

This is a little trickier, but if we look at how we did fmap we can guess at how we get the function and the value out to apply one to the other:

  Cont f <*> Cont x = Cont $ \rest -> f (\fn -> x (\val -> rest (fn val)))

Monad is a harder challenge, but the same basic tactic applies. Hint: remember to unwrap the newtype with runCont, case, or let when necessary.

So what's callCC?

"Call with current continuation". I don't really get the name. Basically, you use callCC like this:

  ret <- callCC $ \exit -> do
    -- A mini Cont block.
    -- You can bind things to ret in one of two ways: either return
    -- something at the end as usual, or call exit with something of
    -- the appropriate type, and the rest of the block will be ignored.
    when (n < 10) $ exit "small!"
    when (n > 100) $ exit "big!"
    return "somewhere in between!"

See if you can work out the type (not too hard: work out the type of exit first, then the do block) then the implementation. Try not to follow the types too much: they will tell you what to write, but not why. Think instead about the strategies we used above, and what each bit means. Hints: remember that exit throws stuff away, and remember to use runCont or similar, as before.

What about ContT?

The thing with ContT is that it's literally exactly the same trick. In fact I think the following definition works fine:

newtype ContT r m a = ContT (Cont (m r) a)
  deriving (Functor, Applicative, Monad)

runContT :: ContT r m a -> (a -> m r) -> m r
runContT (ContT m) = runCont m

The only reason the newtype exists at all is to make the kind compatible with things like MonadTrans.

Some real examples

The examples in the mtl doc are unconvincing. They don't do anything genuinely daring. Some of them work in any monad! Here's a more complex example:

-- This tends to be useful. Don't ask about the name.
reset :: Cont a a -> a
reset c = runCont c id

faff :: Integer -> Maybe Integer
faff n = reset $ do
  test <- Cont $ \try -> case try n of 
    Nothing -> try (2*n) 
    res -> fmap (subtract 10) res
  return $ if test < 10 then Nothing else Just test

The return statement is run with test = n: if it succeeds then we subtract 10 from the result and return it. If it fails we try again, but with (2*n): note that if this succeeds, we don't subtract 10.

As an exercise, work out how to make the function return (a) Nothing, (b) Just 12, (c) Just 0.