# MonadCont under the hood

(Difference between revisions)

This tutorial is a response to the following Stack Overflow question. There's a short but useful description of Cont and MonadCont operations in the Control.Monad.Cont documentation, but it doesn't really describe how the continuation monad does its thing. This is an attempt at much more detailed explanation of what Cont and MonadCont are doing under the hood.

This tutorial assumes a working knowledge of Haskell, though of course it doesn't assume that you understood the implementation of
the first time you read it!

# 1 Introducing Continuations and the Cont type

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 continuation-passing-style 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.

# 2 Sequencing Continuation-Style Computations

## 2.1 Basic 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 invoke
c3
when it's done.
• f2
needs to invoke a continuation
c2
that will invoke
f3
, which will invoke
c3
.
• f1
needs to invoke a continuation
c1
,
which will invoke
f2
,
which will invoke
c2
,
which will invoke
f3
,
which will finally invoke
c3
.
To chain the
Cont
objects together, then, we need to create the appropriate continuation functions
c1
and
c2
and make sure they get passed as the continuation argument to
f1
and
f2
respectively.

## 2.2 Extending to Monad

Extending this idea to the
class in general, 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 a
Cont
object that just passes that value to its continuation.
• The bind operator
(>>=)
takes a
Cont
object, and a function that produces another
Cont
object given a value from the first
, and chains them together into one
Cont
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 with
C

# 3 Understanding the Monad

## 3.1 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." Thus,
return
in the
Cont
monad passes the result of its computation directly to the continuation it's given.

A mathematical analogon that may help is the "insert a" functional: it takes a function f, and the result is the value f(a) of that function at x=a. Call it "insert(a)", then

insert(a)               <--->   return a
insert(a)(f(x) = x^2)   <--->   runCont (return a) (\x -> x^2)
== f(a)                         == (\x -> x^2) a
== a^2                          == a^2

## 3.2 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
by simply passing its value to its continuation. The
Cont
objects are being created "just in time" to be used, based on the computation so far.

# 4 Exploring the Monad

Here's a simple example 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 top-level
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: The sequence of terms
runCont Cont \$
effectively cancel 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:
doC = 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!
If you changed doC to
return 4 >>= h
, the derivation would be almost identical to the above, except that 4 would pass through to h, which would unfold to g instead. "Done: 2" should be the result.

# 5 MonadCont and callCC

One final extension to this monad, which can be extremely useful in practice, is the
class, which provides a
callCC
operation.
callCC
creates a
Cont
object that invokes a function to construct a
Cont
object, and then runs it with the continuation it's given. However, it provides that function an alternate continuation 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 continuation-passing 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.