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1 A practical Cont tutorial

It seems to me like
is way simpler than people make it. I think it's just a way to give a name to the "tail" of a do-block.
contstuff :: Magic a
contstuff = do
  -- Now 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 and combine the results, or discard it entirely,
    -- or do it and then use the result to do it again... it's easy to
    -- imagine why this might be useful.
    messAboutWith rest
  thing3 -- these might get done once, several times,
  thing4 -- or not at all.
The question is, what type should
have? Well, let's say the whole do-block results in a thing of type
(without thinking too hard about what this means). Then certainly the function we give
should result in type
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
, right? Well, more or less, with one caveat: we might bind the result of
  x <- magic $ \rest -> -- ...
  thingInvolving x
so the rest of the do-block has an
in it that we need to supply (as well as other variables, but
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
. So our lambda has type
(a -> r) -> r
and hence
magic :: (a -> r) -> r -> Magic a
Magic a = Cont r a
magic = Cont


The thing with
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
without thinking about the types too much:
instance Functor (Cont r) where
  fmap f (Cont g) = -- ...
Well, we've got to build a
value, and those always start the same way:
  fmap f (Cont g) = Cont $ \rest -> -- ...
Now what? Well, remember what
is. It looks like
\rest -> stuffWith (rest val)
, where
is the 'value' of the computation (what would be bound with
). So we want to give it a
, but we don't want it to be called with the 'value' of the computation - we want
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
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
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)))
is a harder challenge, but the same basic tactic applies. Hint: remember to unwrap the newtype with
, or
when necessary.

1.1 So what's callCC?

"Call with current continuation". I don't really get the name. Basically, you use
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
throws stuff away, and remember to use
or similar, as before.

1.2 What about ContT?

The thing with
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

1.3 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.
runC :: Cont a a -> a
runC c = runCont c id
faff :: Integer -> Maybe Integer
faff n = runC $ 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
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
: note that if this succeeds, we don't subtract 10. As an exercise, work out how to make the function return (a)
, (b)
Just 12
, (c)
Just 0

1.4 Acknowledgements

I think it was sigfpe who made this click for me, after thinking about how this works: Quick and dirty reinversion of control and there's also this: The Mother of all Monads which is more-or-less the above trick but in a bit more detail.