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MonadCont under the hood

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This tutorial is a response to the following [http://stackoverflow.com/questions/3322540/tutorial-to-disassemble-the-haskell-cont-monad/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/Control-Monad-Cont.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 is a response to the following [http://stackoverflow.com/questions/3322540/tutorial-to-disassemble-the-haskell-cont-monad Stack Overflow question]. There's a short but useful description of <tt>Cont</tt> and <tt>MonadCont</tt> operations in the [http://hackage.haskell.org/packages/archive/mtl/1.1.0.2/doc/html/Control-Monad-Cont.html 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 <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!
Line 16: Line 16:
   
 
=Sequencing Continuation-Style Computations=
 
=Sequencing Continuation-Style Computations=
==Applicative Sequencing==
+
==Basic Sequencing==
<hask>Cont</hask> objects can be chained together, so that the continuation you pass in threads through the guts of all the <hask>Cont</hask> objects in the chain before it's finally invoked. The way they chain is the way <hask>Cont</hask> 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 <hask>Cont</hask> objects <hask>F1 -> F2 -> F3</hask>, and let's say you had a continuation <hask>C3</hask> that you want to pass to the chain. Then:
+
<hask>Cont</hask> objects can be chained together, so that the continuation you pass in threads through the guts of all the <hask>Cont</hask> objects in the chain before it's finally invoked. The way they chain is the way <hask>Cont</hask> 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 <hask>Cont</hask> objects <hask>f1 -> f2 -> f3</hask>, and let's say you had a continuation <hask>c3</hask> that you want to pass to the chain. Then:
   
* <hask>F3</hask> needs to invoke <hask>C3</hask>when it's done
+
* <hask>f3</hask> needs to invoke <hask>c3</hask> when it's done.
* <hask>F2</hask> needs to invoke a continuation <hask>C2</hask> that will invoke <hask>F3</hask>, which will invoke <hask>C3</hask>.
+
* <hask>f2</hask> needs to invoke a continuation <hask>c2</hask> that will invoke <hask>f3</hask>, which will invoke <hask>c3</hask>.
* <hask>F1</hask> needs to invoke a continuation <hask>C1</hask> which will invoke <hask>F2</hask>, which will invoke <hask>F3</hask>, which will invoke <hask>C3</hask>.
+
* <hask>f1</hask> needs to invoke a continuation <hask>c1</hask>,<br/>&nbsp;which will invoke <hask>f2</hask>,<br/>&nbsp;&nbsp;which will invoke <hask>c2</hask>,<br/>&nbsp;&nbsp;&nbsp;which will invoke <hask>f3</hask>,<br/>&nbsp;&nbsp;&nbsp;&nbsp;which will finally invoke <hask>c3</hask>.
   
What I've described so far is the ''applicative'' operation of continuations.
+
To chain the <hask>Cont</hask> objects together, then, we need to create the appropriate continuation functions <hask>c1</hask> and <hask>c2</hask> and make sure they get passed as the continuation argument to <hask>f1</hask> and <hask>f2</hask> respectively.
   
 
==Extending to Monad==
 
==Extending to Monad==
With the <hask>Monad</hask> operation there's an extra wrinkle: we allow for the value of one computation to affect ''which'' <hask>Cont</hask> object gets invoked next. In this world:
+
Extending this idea to the <hask>Monad</hask> class in general, there's an extra wrinkle: we allow for the value of one computation to affect ''which'' <hask>Cont</hask> object gets invoked next. In this world:
   
 
* <hask>return</hask> takes a value and produces a <hask>Cont</hask> object that just passes that value to its continuation.
 
* <hask>return</hask> takes a value and produces a <hask>Cont</hask> object that just passes that value to its continuation.
* <hask>bind</hask> takes a <hask>Cont</hask> object, and a ''function that produces another <hask>Cont</hask> object given a value from the first'', and chains them together into one <hask>Cont</hask> object. That object, when invoked, is going to:
+
* The bind operator <hask>(>>=)</hask> takes a <hask>Cont</hask> object, and a ''function that produces another <hask>Cont</hask> object given a value from the first'', and chains them together into one <hask>Cont</hask> object. That object, when invoked, is going to:
 
** take a single continuation object <hask>C</hask>,
 
** take a single continuation object <hask>C</hask>,
 
** produce an intermediate value,
 
** produce an intermediate value,
Line 50: Line 50:
   
 
Why? The code <hask>($ a)</hask> is a ''slice'' of the operator <hask>$</hask>, which represents application. In other words, <hask>($ a)</hask> can be equivalently written <hask>\f -> f a</hask>, or "take a function as input and apply it to a."
 
Why? The code <hask>($ a)</hask> is a ''slice'' of the operator <hask>$</hask>, which represents application. In other words, <hask>($ a)</hask> can be equivalently written <hask>\f -> f a</hask>, or "take a function as input and apply it to a."
  +
  +
Thus, <hask>return</hask> in the <hask>Cont</hask> monad passes the result of its computation directly to the continuation it's given.
   
 
==Bind==
 
==Bind==
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</haskell>
 
</haskell>
   
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 <hask>Cont</hask> objects are being created "just in time" to be used, based on the computation so far.
+
Do you see what's happening? <hask>(k a)</hask> has become part of the continuation that <hask>m</hask> is given, and <hask>m</hask> passes its value to <hask>k</hask> by simply passing its value to its continuation. The <hask>Cont</hask> objects are being created "just in time" to be used, based on the computation so far.
   
=Applying the Monad=
+
=Exploring the Monad=
Here's a simple example I've cooked up that should help illustrate the monad in action:
+
Here's a simple example that should help illustrate the monad in action:
   
 
<haskell>
 
<haskell>
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<hask>h</hask> is a function that takes a value and produces a <hask>Cont</hask> object depending on the value it's given.
 
<hask>h</hask> is a function that takes a value and produces a <hask>Cont</hask> object depending on the value it's given.
   
''Lemma:'' <hask>runCont Cont $</hask> effectively cancels out, i.e. <hask>runCont (Cont $ \c -> ...)</hask> is simply the function <hask>\c -> ...</hask>. This is because <hask>runCont</hask> is a field selector of <hask>Cont</hask> objects, and <hask>Cont</hask> objects only have that one field.
+
''Lemma:'' The sequence of terms <hask>runCont Cont $</hask> effectively cancel out, i.e. <hask>runCont (Cont $ \c -> ...)</hask> is simply the function <hask>\c -> ...</hask>. This is because <hask>runCont</hask> is a field selector of <hask>Cont</hask> objects, and <hask>Cont</hask> objects only have that one field.
   
 
Therefore, <hask>(return 5) >>= h</hask> expands and simplifies to:
 
Therefore, <hask>(return 5) >>= h</hask> expands and simplifies to:
   
 
<haskell>
 
<haskell>
let s c = c 5
+
doC = let s c = c 5
t c = \a -> runCont (h a) c
+
t c = \a -> runCont (h a) c
in Cont $ \c -> s (t c)
+
in Cont $ \c -> s (t c)
 
</haskell>
 
</haskell>
   
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=> "Done: 15" -- apply *; apply finalC to final value!
 
=> "Done: 15" -- apply *; apply finalC to final value!
   
=Understanding MonadCont and callCC=
+
If you changed doC to <hask>return 4 >>= h</hask>, 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.
One final extension to this which is frequently used is the <hask>MonadCont</hask> class, which provides a <hask>callCC</hask> operation. <hask>callCC</hask> creates a <hask>Cont</hask> 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 <hask>callCC</hask> 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:
+
  +
=MonadCont and callCC=
  +
One final extension to this monad, which can be extremely useful in practice, is the <hask>MonadCont</hask> class, which provides a <hask>callCC</hask> operation. <hask>callCC</hask> creates a <hask>Cont</hask> object that invokes a function to construct a <hask>Cont</hask> 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 <hask>callCC</hask> 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:
   
 
<haskell>
 
<haskell>
Line 185: Line 187:
 
</haskell>
 
</haskell>
   
The key is <hask>backtrack</hask>, which takes whatever "inner" continuation is active when backtrack is invoked, completely ignores it, and simply passes its value to the "outer" continuation <hask>c</hask>. (Compare this to the definition of <hask>return</hask>, which always uses the continuation it's given.) <hask>f</hask>is the function passed to <hask>callCC</hask>, whose extent provides the context under which <hask>backtrack</hask> can be used.
+
The key is <hask>backtrack</hask>, which takes whatever "inner" continuation is active when backtrack is invoked, completely ignores it, and simply passes its value to the "outer" continuation <hask>c</hask>. (Compare this to the definition of <hask>return</hask>, which always uses the continuation it's given.) <hask>f</hask> is the function passed to <hask>callCC</hask>, whose extent provides the context under which <hask>backtrack</hask> can be used.
   
 
[[Category:Monad]]
 
[[Category:Monad]]

Revision as of 00:29, 26 July 2010

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
Control.Monad.Cont
the first time you read it!

Contents

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
Monad
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.

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
MonadCont
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.