Difference between revisions of "MonadCont under the hood"
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−  This tutorial is a response to the following [http://stackoverflow.com/questions/3322540/tutorialtodisassemblethehaskellcontmonad/3323283#3323283 Stack Overflow question]. There's a short but useful description of Cont and MonadCont in the Control.Monad.Cont documentation, but it doesn't really describe how the continuation monad works. This is an attempt at much more detailed explanation of what Cont and MonadCont are doing, particularly below the hood. 
+  This tutorial is a response to the following [http://stackoverflow.com/questions/3322540/tutorialtodisassemblethehaskellcontmonad/3323283#3323283 Stack Overflow question]. There's a short but useful description of <tt>Cont</tt> and <tt>MonadCont</tt> in the [http://hackage.haskell.org/packages/archive/mtl/1.1.0.2/doc/html/ControlMonadCont.html Control.Monad.Cont documentation], but it doesn't really describe how the continuation monad works. This is an attempt at much more detailed explanation of what Cont and MonadCont are doing, particularly below the hood. 
This tutorial assumes a working knowledge of Haskell, though of course it doesn't assume that you understood the implementation of <hask>Control.Monad.Cont</hask> the first time you read it! 
This tutorial assumes a working knowledge of Haskell, though of course it doesn't assume that you understood the implementation of <hask>Control.Monad.Cont</hask> the first time you read it! 
Revision as of 02:41, 24 July 2010
This tutorial is a response to the following Stack Overflow question. There's a short but useful description of Cont and MonadCont in the Control.Monad.Cont documentation, but it doesn't really describe how the continuation monad works. This is an attempt at much more detailed explanation of what Cont and MonadCont are doing, particularly below the hood.
This tutorial assumes a working knowledge of Haskell, though of course it doesn't assume that you understood the implementation of Control.Monad.Cont
the first time you read it!
Contents
Continuations and the Cont monad
Continuations are functions that represent "the remaining computation to do." Their representation here is a > r
, which is simply a function that takes some value produced by the current computation, of some type a
, and returns the final result of type r
from it.
The type Cont r a
(instances of which I will, in this tutorial, refer to as Cont objects) represents a continuationpassingstyle function that takes a single continuation as its only input. In other words, its guts are a function that:
1. takes a continuation as an argument
2. does whatever it needs to do
3. produces a value of type r
at the end, presumably by invoking the continuation.
Note that whatever it needs to do, i.e. whatever values it needs to be able to use to do its thing, must already be bound up into the Cont
object. So, generally, we won't be dealing with Cont
objects directly, but with functions that can ultimately produce one.
Sequencing ContinuationStyle Computations
Applicative Sequencing
Cont
objects can be chained together, so that the continuation you pass in threads through the guts of all the Cont
objects in the chain before it's finally invoked. The way they chain is the way Cont
works: each object in the chain invokes a continuation that has the next object's computation prepended to the final continuation. Let's say we have a chain of Cont
objects F1 > F2 > F3
, and let's say you had a continuation C3
that you want to pass to the chain. Then:

F3
needs to invokeC3
when it's done 
F2
needs to invoke a continuationC2
that will invokeF3

F1
needs to invoke a continuationC1
that will invokeF2
, thenF3
, thenC3
.
What I've described so far is the Applicative
operation of continuations.
Extending to Monad
With the Monad
operation there's an extra wrinkle: we allow for the value of one computation to affect which Cont
object gets invoked next. In this world:

return
takes a value and produces aCont
object that just passes that value to its continuation. 
bind
takes aCont
object, and a function that produces anotherCont
object given a value from the first, and chains them together into oneCont
object. That object, when invoked, is going to: take a single continuation object
C
,  produce an intermediate value,
 use that intermediate value to select/create the next
Cont
object to invoke,  invoke that
Cont
object withC
 take a single continuation object
Understanding the Monad
Return
The code:
return a = Cont ($ a)
is equivalent to the following code:
return a = Cont $ \c > c a
Why? The code ($ a)
is a slice of the operator $
, which represents application. In other words, ($ a)
can be equivalently written \f > f a
, or "take a function as input and apply it to a."
Bind
The code:
m >>= k = Cont $ \c > runCont m $ \a > runCont (k a) c
is a terse way of saying the following:
m >>= k = let s c = runCont m c
t c = \a > runCont (k a) c
in Cont $ \c > s (t c)
Do you see what's happening? (k a) has become part of the continuation that m is given, and m passes its value to k simply by passing its value to its continuation. The Cont
objects are being created "just in time" to be used, based on the computation so far.
Applying the Monad
Here's a simple example I've cooked up that should help illustrate the monad in action:
f :: Int > Cont r Int
f x = Cont $ \c > c (x * 3)
g :: Int > Cont r Int
g x = Cont $ \c > c (x  2)
These are simple functions that produce Cont
objects, given an intermediate value x
. You can see that the value being passed to the continuation is an Int, though we don't put any restrictions on what that continuation can ultimately produce.
BTW, they can be equivalently written as:
f x = return (x * 3)
g x = return (x  2)
where they look very similar to normal functions. I'm writing them longhand to show you explicitly what the functions are doing.
h :: Int > Cont r Int
h x  x == 5 = f x
 otherwise = g x
This is a more complicated function that chooses between two other Cont
objects, based on the input it's given.
Now let's create a toplevel Cont
object that does some chaining:
doC :: Cont r Int
doC = return 5 >>= h
And we'll invoke it like this:
finalC :: Show a => a > String
finalC x = "Done: " ++ show(x)
runCont doC finalC
Note that runCont doC
produces a function of type (Int > a) > a
, which is invoked on a continuation of type Show a => a > String
, which reduces in this context to Int > String
. The final value produced will be a String
. Can you guess what it will say? What if you changed return 5
to return 4
?
Let's see if you're right:
return 5
produces a Cont
object that basically looks like this: Cont $ \c > c 5
. So that part is easy.
h
is a function that takes a value and produces a Cont
object depending on the value it's given.
Lemma: runCont Cont $
effectively cancels out, i.e. runCont (Cont $ \c > ...)
is simply the function \c > ...
. This is because runCont
is a field selector of Cont
objects, and Cont
objects only have that one field.
Therefore, (return 5) >>= h
expands and simplifies to:
let s c = c 5
t c = \a > runCont (h a) c
in Cont $ \c > s (t c)
And finally, runCont doC finalC
evaluates to:
runCont doC finalC => runCont (Cont $ \c > s (t c)) finalC  unfold doC => s (t finalC)  simplify with lemma and apply to finalC => (t finalC) 5  unfold s => (\a > runCont (h a) finalC) 5  unfold t => runCont (h 5) finalC  apply \a... to 5 => runCont (f 5) finalC  unfold h => runCont (Cont $ \c > c (5*3)) finalC  unfold f => (\c > c (5*3)) finalC  simplify with lemma => finalC (5*3)  apply \c... to finalC => "Done: 15"  apply *; apply finalC to final value!
Understanding MonadCont and callCC
One final extension to this which is frequently used is the MonadCont
class, which provides a callCC
operation. callCC
creates a Cont
object that invokes the function it's given, but provides a second continuation to that function that can be invoked to "break out" of the computation and simply pass a value to the continuation that was active when callCC
was invoked. This function's operation is definitely easier to understand by seeing it in action. Evaluate the following code, replacing the corresponding functions above:
h :: Int > (Int > Cont r Int) > Cont r Int
h x abort  x == 5 = f x
 otherwise = abort (1)
doC n = return n >>= \x >
callCC (\abort > h x abort) >>= \y >
g y
Run runCont (doC 5) finalC
. h should invoke f, and g will be invoked afterward, so you should get 13 as the final answer.
Now change (doC 5)
to (doC 4)
. In this case, h will call abort, which passes 1 to g. 3 should be the final answer.
Now change doC
to move g inside the callCC abort context:
doC n = return n >>= \x >
callCC (\abort > h x abort >>= \y >
g y)
and run with (doC 4)
. In this case, h invokes abort and g is never invoked! 1 is the final answer.
Once you've converted all your operations to continuationpassing style by putting them in the Cont
monad, and have a handle on how >>=
works in that monad, understanding how callCC
works is surprisingly simple:
callCC f = Cont $ \c > runCont (f (\a > Cont $ \_ > c a )) c
can be written as
callCC f = let backtrack a = Cont $ \_ > c a
in Cont $ \c > runCont (f backtrack) c
The key is backtrack
, which takes whatever "inner" continuation is active when backtrack is invoked, completely ignores it, and simply passes its value to the "outer" continuation c
. (Compare this to the definition of return
, which always uses the continuation it's given.) f
is the function passed to callCC
, whose extent provides the context under which backtrack
can be used.