Refunctionalizing zippers: from lists to continuations
If zippers are continuations reified (defuntionalized), then one route to continuations is to re-functionalize a zipper. Then the concreteness and understandability of the zipper provides a way of understanding an equivalent treatment using continuations.
Let's work with lists of
chars for a change. We'll sometimes write
"abSd" as an abbreviation for
['a'; 'b'; 'S'; 'd'].
We will set out to compute a deceptively simple-seeming task: given a string, replace each occurrence of 'S' in that string with a copy of the string up to that point.
We'll define a function
t (for "task") that maps strings to their
t "abSd" ~~> "ababd"
In linguistic terms, this is a kind of anaphora
'S' is functioning like an anaphoric element, and
the preceding string portion is the antecedent.
This task can give rise to considerable complexity. Note that it matters which 'S' you target first (the position of the * indicates the targeted 'S'):
t "aSbS" * ~~> t "aabS" * ~~> "aabaab"
t "aSbS" * ~~> t "aSbaSb" * ~~> t "aabaSb" * ~~> "aabaaabab"
t "aSbS" * ~~> t "aSbaSb" * ~~> t "aSbaaSbab" * ~~> t "aSbaaaSbaabab" * ~~> ...
Apparently, this task, as simple as it is, is a form of computation,
and the order in which the
'S's get evaluated can lead to divergent
For now, we'll agree to always evaluate the leftmost
guarantees termination, and a final string without any
'S' in it.
This is a task well-suited to using a zipper. We'll define a function
tz (for task with zippers), which accomplishes the task by mapping a
char list zipper to a
char list. We'll call the two parts of the
zipped; we start with a fully zipped list, and
move elements to the unzipped part by pulling the zipper down until the
entire list has been unzipped, at which point the zipped half of the
zipper will be empty.
type 'a list_zipper = ('a list) * ('a list);; let rec tz (z : char list_zipper) = match z with | (unzipped, ) -> List.rev(unzipped) (* Done! *) | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) # tz (, ['a'; 'b'; 'S'; 'd']);; - : char list = ['a'; 'b'; 'a'; 'b'; 'd'] # tz (, ['a'; 'S'; 'b'; 'S']);; - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
Note that the direction in which the zipper unzips enforces the evaluate-leftmost rule. Task completed.
One way to see exactly what is going on is to watch the zipper in
action by tracing the execution of
tz. By using the
directive in the OCaml interpreter, the system will print out the
tz each time it is called, including when it is called
recursively within one of the
match clauses. Note that the
lines with left-facing arrows (
<--) show (both initial and recursive) calls to
giving the value of its argument (a zipper), and the lines with
right-facing arrows (
-->) show the output of each recursive call, a
# #trace tz;; t1 is now traced. # tz (, ['a'; 'b'; 'S'; 'd']);; tz <-- (, ['a'; 'b'; 'S'; 'd']) (* Initial call *) tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *) tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *) tz <-- (['b'; 'a'; 'b'; 'a'], ['d']) (* Special 'S' step *) tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], ) (* Pull zipper *) tz --> ['a'; 'b'; 'a'; 'b'; 'd'] (* Output reversed *) tz --> ['a'; 'b'; 'a'; 'b'; 'd'] tz --> ['a'; 'b'; 'a'; 'b'; 'd'] tz --> ['a'; 'b'; 'a'; 'b'; 'd'] tz --> ['a'; 'b'; 'a'; 'b'; 'd'] - : char list = ['a'; 'b'; 'a'; 'b'; 'd']
The nice thing about computations involving lists is that it's so easy to visualize them as a data structure. Eventually, we want to get to a place where we can talk about more abstract computations. In order to get there, we'll first do the exact same thing we just did with concrete zipper using procedures instead.
Think of a list as a procedural recipe:
['a'; 'b'; 'c'; 'd'] is the result of
'a'::('b'::('c'::('d'::))) (or, in our old style,
make_list 'a' (make_list 'b' (make_list 'c' (make_list 'd' empty)))). The
recipe for constructing the list goes like this:
(0) Start with the empty list 
(1) make a new list whose first element is 'd' and whose tail is the list constructed in step (0)
(2) make a new list whose first element is 'c' and whose tail is the list constructed in step (1)
(3) make a new list whose first element is 'b' and whose tail is the list constructed in step (2)
(4) make a new list whose first element is 'a' and whose tail is the list constructed in step (3)
What is the type of each of these steps? Well, it will be a function
from the result of the previous step (a list) to a new list: it will
be a function of type
char list -> char list. We'll call each step
(or group of steps) a continuation of the previous steps. So in this
context, a continuation is a function of type
char list -> char
list. For instance, the continuation corresponding to the portion of
the recipe below the horizontal line is the function
fun (tail : char
list) -> 'a'::('b'::tail). What is the continuation of the 4th step? That is, after we've built up
'a'::('b'::('c'::('d'::))), what more has to happen to that for it to become the list
['a'; 'b'; 'c'; 'd']? Nothing! Its continuation is the function that does nothing:
fun tail -> tail.
In what follows, we'll be thinking about the result list that we're building up in this procedural way. We'll treat our input list just as a plain old static list data structure, that we recurse through in the normal way we're accustomed to. We won't need a zipper data structure, because the continuation-based representation of our result list will take over the same role.
So our new function
tc (for task with continuations) takes an input list (not a zipper) and a also takes a continuation
k (it's conventional to use
k for continuation variables).
k is a function that represents how the result list is going to continue being built up after this invocation of
tc delivers up a value. When we invoke
tc for the first time, we expect it to deliver as a value the very de-S'd list we're seeking, so the way for the list to continue being built up is for nothing to happen to it. That is, our initial invocation of
tc will supply
fun tail -> tail as the value for
k. Here is the whole
tc function. Its structure and behavior follows
tz from above, which we've repeated here to facilitate detailed comparison:
let rec tz (z : char list_zipper) = match z with | (unzipped, ) -> List.rev(unzipped) (* Done! *) | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) let rec tc (l: char list) (k: (char list) -> (char list)) = match l with |  -> List.rev (k ) | 'S'::zipped -> tc zipped (fun tail -> k (k tail)) | target::zipped -> tc zipped (fun tail -> target::(k tail));; # tc ['a'; 'b'; 'S'; 'd'] (fun tail -> tail);; - : char list = ['a'; 'b'; 'a'; 'b'] # tc ['a'; 'S'; 'b'; 'S'] (fun tail -> tail);; - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
To emphasize the parallel, we've re-used the names
target. The trace of the procedure will show that these variables
take on the same values in the same series of steps as they did during
the execution of
tz above: there will once again be one initial and
four recursive calls to
zipped will take on the values
"" (and, once again, on the final call,
match clause will fire, so the the variable
not be instantiated).
We have not named the continuation argument
unzipped, although that is
what the parallel would suggest. The reason is that
tz) is a list, but
tc) is a function. That's the most crucial
difference between the solutions---it's the
point of the excercise, and it should be emphasized. For instance,
you can see this difference in the fact that in
tz, we have to glue
together the two instances of
unzipped with an explicit (and,
computationally speaking, relatively inefficient)
tc version of the task, we simply compose
k with itself:
k o k = fun tail -> k (k tail).
tc ['a'; 'b'; 'S'; 'd'] would yield a partially-applied function; it would still wait for another argument, a continuation of type
char list -> char list. So we have to give it an "initial continuation" to get started. As mentioned above, we supply the identity function as the initial continuation. Why did we choose that? Again, if
you have already constructed the result list
"ababd", what's the desired continuation? What's the next step in the recipe to produce the desired result, i.e, the very same list,
"ababd"? Clearly, the identity function.
A good way to test your understanding is to figure out what the
k must be at the point in the computation when
tc is applied to the argument
"Sd". Two choices: is it
fun tail -> 'a'::'b'::tail, or it is
fun tail -> 'b'::'a'::tail? The way to see if you're right is to execute the following command and see what happens:
tc ['S'; 'd'] (fun tail -> 'a'::'b'::tail);;
There are a number of interesting directions we can go with this task.
The reason this task was chosen is because the task itself (as opposed
to the functions used to implement the task) can be viewed as a
simplified picture of a computation using continuations, where
plays the role of a continuation operator. (It works like the Scheme
control; the differences between them don't
manifest themselves in this example.
See Ken Shan's paper Shift to control,
which inspired some of the discussion in this topic.)
In the analogy, the input list portrays a
sequence of functional applications, where
[f1; f2; f3; x] represents
f1(f2(f3 x)). The limitation of the analogy is that it is only
possible to represent computations in which the applications are
always right-branching, i.e., the computation
((f1 f2) f3) x cannot
be directly represented.
One way to extend this exercise would be to add a special symbol
and then the task would be to copy from the target
'S' only back to
'#'. This would allow our task to simulate delimited
continuations with embedded
prompts (also called
Here is some Scheme code implementing the
tc functions, first as presented above, and second with the variant just mentioned, using
'#'. There's also a third kind of implementation, which is akin to the
tc version, but doesn't explicitly pass a
k argument, and instead uses these unfamiliar operations
shift. We'll be explaining what these do shortly. (The reason this code is in Scheme is because that's the language in which it's easiest to work with operations like