From 67cfadf1ab1d26bfa2e40ba5a675fbcdd7ea5e01 Mon Sep 17 00:00:00 2001 From: Chris Barker Date: Wed, 1 Dec 2010 11:08:57 -0500 Subject: [PATCH] edits --- from_list_zippers_to_continuations.mdwn | 38 +++++++++++++++++++++------------ 1 file changed, 24 insertions(+), 14 deletions(-) diff --git a/from_list_zippers_to_continuations.mdwn b/from_list_zippers_to_continuations.mdwn index 1ef5f9c0..8b1fa684 100644 --- a/from_list_zippers_to_continuations.mdwn +++ b/from_list_zippers_to_continuations.mdwn @@ -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'] @@ -170,15 +172,17 @@ 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 functional 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, @@ -191,16 +195,22 @@ you have already constructed the initial list `"abSd"`, what's the desired conti 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 +`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 -- 2.11.0