X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=from_list_zippers_to_continuations.mdwn;h=dcd11cec2181b4b98f37477e0fbd693ba889a7fa;hp=1ef5f9c065fc670a15e571c3b36eb2ba15092d77;hb=b62e35e77aaba7acf2441ce120931e329f61b774;hpb=0feebbaaa58403d836d7ea6166cf709dd3faf1a8 diff --git a/from_list_zippers_to_continuations.mdwn b/from_list_zippers_to_continuations.mdwn index 1ef5f9c0..dcd11cec 100644 --- a/from_list_zippers_to_continuations.mdwn +++ b/from_list_zippers_to_continuations.mdwn @@ -6,9 +6,9 @@ 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 `char`s for a change. To maximize readability, we'll -indulge in an abbreviatory convention that "abSd" abbreviates the -list `['a'; 'b'; 'S'; 'd']`. +Let's work with lists of `char`s 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 @@ -70,7 +70,8 @@ This is a task well-suited to using a zipper. We'll define a function `char list zipper` to a `char list`. We'll call the two parts of the zipper `unzipped` and `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 (and so the zipped half of the zipper is empty). +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);; @@ -86,14 +87,15 @@ entire list has been unzipped (and so the zipped half of the zipper is empty). # tz ([], ['a'; 'S'; 'b'; 'S']);; - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -Note that this implementation enforces the evaluate-leftmost rule. -Task completed. +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 `#trace` directive in the OCaml interpreter, the system will print out the -arguments to `tz` each time it is (recursively) called. Note that the -lines with left-facing arrows (`<--`) show (recursive) calls to `tz`, +arguments to `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 `tz`, giving the value of its argument (a zipper), and the lines with right-facing arrows (`-->`) show the output of each recursive call, a simple list. @@ -101,10 +103,10 @@ simple list. # #trace tz;; t1 is now traced. # tz ([], ['a'; 'b'; 'S'; 'd']);; - 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 step *) + 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'] @@ -119,14 +121,14 @@ 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'; 'S'; 'd']` is the result of -the computation `'a'::('b'::('S'::('d'::[])))` (or, in our old style, -`make_list 'a' (make_list 'b' (make_list 'S' (make_list 'd' empty)))`). The +Think of a list as a procedural recipe: `['a'; 'b'; 'c'; 'd']` is the result of +the computation `'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 'S' and whose tail is the list constructed in step (1) +> (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) @@ -138,16 +140,11 @@ be a function of type `char list -> char list`. We'll call each step 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)`. +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`. -This means that we can now represent the unzipped part of our -zipper as a continuation: a function -describing how to finish building a list. We'll write a new -function, `tc` (for task with continuations), that will take an input -list (not a zipper!) and a continuation `k` (it's conventional to use `k` for continuation variables) and return a processed list. -The structure and the behavior will follow that of `tz` above, with -some small but interesting differences. We've included the orginal -`tz` to facilitate detailed comparison: +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 @@ -170,15 +167,16 @@ some small but interesting differences. We've included the orginal To emphasize the parallel, we've re-used the names `zipped` and `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 +the execution of `tz` above: there will once again be one initial and four recursive calls to `tc`, and `zipped` will take on the values `"bSd"`, `"Sd"`, `"d"`, and `""` (and, once again, on the final call, the first `match` clause will fire, so the the variable `zipped` will not be instantiated). -We have not called the functional argument `unzipped`, although that is -what the parallel would suggest. The reason is that `unzipped` is a -list, but `k` is a function. That's the most crucial difference, the +We have not named the continuation argument `unzipped`, although that is +what the parallel would suggest. The reason is that `unzipped` (in +`tz`) is a list, but `k` (in `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, @@ -186,21 +184,26 @@ computationally speaking, relatively inefficient) `List.append`. In the `tc` version of the task, we simply compose `k` with itself: `k o k = fun tail -> k (k tail)`. -A call `tc ['a'; 'b'; 'S'; 'd']` yields a partially-applied function; it still waits for another argument, a continuation of type `char list -> char list`. We have to give it an "initial continuation" to get started. Here we supply *the identity function* as the initial continuation. Why did we choose that? Well, if -you have already constructed the initial list `"abSd"`, what's the desired continuation? What's the next step in the recipe to produce the desired result, i.e, the very same list, `"abSd"`? Clearly, the identity function. +A call `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 continuation function `k` must be at the point in the computation when -`tc` is called with the first 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` 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 it can be viewed as a +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 `'S'` -plays the role of a continuation operator. (It works like the Scheme operators `shift` or `control`; the differences between them don't manifest themselves in this example.) In the analogy, the input list portrays a +plays the role of a continuation operator. (It works like the Scheme +operators `shift` or `control`; the differences between them don't +manifest themselves in this example. +See Ken Shan's paper [Shift to control](http://www.cs.rutgers.edu/~ccshan/recur/recur.pdf), +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 @@ -213,7 +216,7 @@ the closest `'#'`. This would allow our task to simulate delimited continuations with embedded `prompt`s (also called `reset`s). The reason the task is well-suited to the list zipper is in part -because the list monad has an intimate connection with continuations. +because the List monad has an intimate connection with continuations. We'll explore this next.