- (define gamma
- (lambda (a lst)
- (letrec ([aux (lambda (l k)
- (cond
- [(null? l) (k 'notfound)]
- [(eq? (car l) a) (cdr l)]
- [(atom? (car l)) (cons (car l) (aux (cdr l) k))]
- [else
- ; what happens when (car l) exists but isn't an atom?
- (let ([car2 (let/cc k2 ; now what will happen when k2 is called?
- (aux (car l) k2))])
- (cond
- ; when will the following condition be met? what happens then?
- [(eq? car2 'notfound) (cons (car l) (aux (cdr l) k))]
- [else (cons car2 (cdr l))]))]))]
- [lst2 (let/cc k1 ; now what will happen when k1 is called?
- (aux lst k1))])
- (cond
- ; when will the following condition be met?
- [(eq? lst2 'notfound) lst]
- [else lst2]))))
-
-Here is [the answer](/hints/cps_hint_3), but try to figure it out for yourself.
-
-Here is the hardest example. Try to figure out what this function does:
-
- (define delta
- (letrec ([yield (lambda (x) x)]
- [resume (lambda (x) x)]
- [walk (lambda (l)
- (cond
- ; is this the only case where walk returns a non-atom?
- [(null? l) '()]
- [(atom? (car l)) (begin
- (let/cc k2 (begin
- (set! resume k2) ; now what will happen when resume is called?
- ; when the next line is executed, what will yield be bound to?
- (yield (car l))))
- ; when will the next line be executed?
- (walk (cdr l)))]
- [else (begin
- ; what will the value of the next line be? why is it ignored?
- (walk (car l))
- (walk (cdr l)))]))]
- [next (lambda () ; next is a thunk
- (let/cc k3 (begin
- (set! yield k3) ; now what will happen when yield is called?
- ; when the next line is executed, what will resume be bound to?
- (resume 'blah))))]
- [check (lambda (prev)
- (let ([n (next)])
- (cond
- [(eq? n prev) #t]
- [(atom? n) (check n)]
- ; when will n fail to be an atom?
- [else #f])))])
- (lambda (lst)
- (let ([fst (let/cc k1 (begin
- (set! yield k1) ; now what will happen when yield is called?
- (walk lst)
- ; when will the next line be executed?
- (yield '())))])
- (cond
- [(atom? fst) (check fst)]
- ; when will fst fail to be an atom?
- [else #f])
- ))))
-
-Here is [the answer](/hints/cps_hint_4), but again, first try to figure it out for yourself.
-
-
-Delimited control operators
-===========================
-
-Here again is the CPS transform for `callcc`:
-
- [callcc (\k. body)] = \outk. (\k. [body] outk) (\v localk. outk v)
-
-`callcc` is what's known as an *undelimited control operator*. That is, the continuations `outk` that get bound into our `k`s include all the code from the `call/cc ...` out to *and including* the end of the program. Calling such a continuation will never return any value to the call site.
-
-(See the technique employed in the `delta` example above, with the `(begin (let/cc k2 ...) ...)`, for a work-around. Also. if you've got a copy of *The Seasoned Schemer*, see the comparison of let/cc vs. "collector-using" (that is, partly CPS) functions at pp. 155-164.)
-
-Often times it's more useful to use a different pattern, where we instead capture only the code from the invocation of our control operator out to a certain boundary, not including the end of the program. These are called *delimited control operators*. A variety of these have been formulated. The most well-behaved from where we're coming from is the pair `reset` and `shift`. `reset` sets the boundary, and `shift` binds the continuation from the position where it's invoked out to that boundary.
-
-It works like this:
-
- (1) outer code
- ------- reset -------
- | (2) |
- | +----shift k ---+ |
- | | (3) | |
- | | | |
- | | | |
- | +---------------+ |
- | (4) |
- +-------------------+
- (5) more outer code
-
-First, the code in position (1) runs. Ignore position (2) for the moment. When we hit the `shift k`, the continuation between the `shift` and the `reset` will be captured and bound to `k`. Then the code in position (3) will run, with `k` so bound. The code in position (4) will never run, unless it's invoked through `k`. If the code in position (3) just ends with a regular value, and doesn't apply `k`, then the value returned by (3) is passed to (5) and the computation continues.
-
-So it's as though the middle box---the (2) and (4) region---is by default not evaluated. This code is instead bound to `k`, and it's up to other code whether and when to apply `k` to any argument. If `k` is applied to an argument, then what happens? Well it will be as if that were the argument supplied by (3) only now that argument does go to the code (4) syntactically enclosing (3). When (4) is finished, that value also goes to (5) (just as (3)'s value did when it ended with a regular value). `k` can be applied repeatedly, and every time the computation will traverse that same path from (4) into (5).
-
-I set (2) aside a moment ago. The story we just told is a bit too simple because the code in (2) needs to be evaluated because some of it may be relied on in (3).
-
-For instance, in Scheme this:
-
- (require racket/control)
-
- (reset
- (let ([x 1])
- (+ 10 (shift k x))))
-
-will return 1. The `(let ([x 1]) ...` part is evaluated, but the `(+ 10 ...` part is not.
-
-Notice we had to preface the Scheme code with `(require racket/control)`. You don't have to do anything special to use `call/cc` or `let/cc`; but to use the other control operators we'll discuss you do have to include that preface in Racket.
-
-This pattern should look somewhat familiar. Recall from our discussion of aborts, and repeated at the top of this page:
-
- let foo x =
- +---try begin----------------+
- | (if x = 1 then 10 |
- | else abort 20 |
- | ) + 100 |
- +---end----------------------+
- in (foo 2) + 1000;;
-
-The box is working like a reset. The `abort` is implemented with a `shift`. Earlier, we refactored our code into a more CPS form:
-
- let x = 2
- in let snapshot = fun box ->
- let foo_result = box
- in (foo_result) + 1000
- in let continue_normally = fun from_value ->
- let value = from_value + 100
- in snapshot value
- in
- if x = 1 then continue_normally 10
- else snapshot 20;;
-
-`snapshot` here corresponds to the code outside the `reset`. `continue_normally` is the middle block of code, between the `shift` and its surrounding `reset`. This is what gets bound to the `k` in our `shift`. The `if...` statement is inside a `shift`. Notice there that we invoke the bound continuation to "continue normally". We just invoke the outer continuation, saved in `snapshot` when we placed the `reset`, to skip the "continue normally" code and immediately abort to outside the box.
-
----
-
-Examples of shift/reset/abort
------------------------------
-
-Here are some more examples of using delimited control operators. We present each example three ways: first a Scheme formulation; then we compute the same result using CPS and the lambda evaluator; then we do the same using the Continuation monad in OCaml. (We don't demonstrate the use of Oleg's delimcc library.)
-
-
-Example 1:
-
- ; (+ 1000 (+ 100 (abort 11))) ~~> 11
-
- app2 (op2 plus) (var thousand)
- (app2 (op2 plus) (var hundred) (abort (var eleven)))
-
- # Continuation_monad.(run0(
- abort 11 >>= fun i ->
- unit (100+i) >>= fun j ->
- unit (1000+j)));;
- - : int = 11
-
-When no `reset` is specified, there's understood to be an implicit one surrounding the entire computation (but unlike in the case of `callcc`, you still can't capture up to *and including* the end of the computation). So it makes no difference if we say instead:
-
- # Continuation_monad.(run0(
- reset (
- abort 11 >>= fun i ->
- unit (100+i) >>= fun j ->
- unit (1000+j))));;
- - : int = 11
-
-
-Example 2:
-
- ; (+ 1000 (reset (+ 100 (abort 11)))) ~~> 1011
-
- app2 (op2 plus) (var thousand)
- (reset (app2 (op2 plus) (var hundred) (abort (var eleven))))
-
- # Continuation_monad.(run0(
- reset (
- abort 11 >>= fun i ->
- unit (100+i)
- ) >>= fun j ->
- unit (1000+j)));;
- - : int = 1011
-
-Example 3:
-
- ; (+ 1000 (reset (+ 100 (shift k (+ 10 1))))) ~~> 1011
-
- app2 (op2 plus) (var thousand)
- (reset (app2 (op2 plus) (var hundred)
- (shift (\k. ((op2 plus) (var ten) (var one))))))
-
- Continuation_monad.(
- let v = reset (
- let u = shift (fun k -> unit (10 + 1))
- in u >>= fun x -> unit (100 + x)
- ) in let w = v >>= fun x -> unit (1000 + x)
- in run0 w);;
- - : int = 1011
-
-Example 4:
-
- ; (+ 1000 (reset (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111
-
- app2 (op2 plus) (var thousand)
- (reset (app2 (op2 plus) (var hundred)
- (shift (\k. (app (var k) ((op2 plus) (var ten) (var one)))))))
-
- Continuation_monad.(
- let v = reset (
- let u = shift (fun k -> k (10 :: [1]))
- in u >>= fun x -> unit (100 :: x)
- ) in let w = v >>= fun x -> unit (1000 :: x)
- in run0 w);;
- - : int list = [1000; 100; 10; 1]
-
-To demonstrate the different adding order between Examples 4 and 5, we use `::` in the OCaml version instead of `+`. Here is Example 5:
-
- ; (+ 1000 (reset (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently
-
- app2 (op2 plus) (var thousand)
- (reset (app2 (op2 plus) (var hundred)
- (shift (\k. ((op2 plus) (var ten) (app (var k) (var one)))))))
-
- Continuation_monad.(let v = reset (
- let u = shift (fun k -> k [1] >>= fun x -> unit (10 :: x))
- in u >>= fun x -> unit (100 :: x)
- ) in let w = v >>= fun x -> unit (1000 :: x)
- in run0 w);;
- - : int list = [1000; 10; 100; 1]
-
-
-Example 6:
-
- ; (+ 100 ((reset (+ 10 (shift k k))) 1)) ~~> 111
-
- app2 (op2 plus) (var hundred)
- (app (reset (app2 (op2 plus) (var ten)
- (shift (\k. (var k))))) (var one))
-
- (* not sure if this example can be typed as-is in OCaml... this is the best I an do at the moment... *)
-
- # type 'x either = Left of (int -> ('x,'x either) Continuation_monad.m) | Right of int;;
- # Continuation_monad.(let v = reset (
- shift (fun k -> unit (Left k)) >>= fun i -> unit (Right (10+i))
- ) in let w = v >>= fun (Left k) ->
- k 1 >>= fun (Right i) ->
- unit (100+i)
- in run0 w);;
- - : int = 111
-
-<!--
-# type either = Left of (int -> either) | Right of int;;
-# let getleft e = match e with Left lft -> lft | Right _ -> failwith "not a Left";;
-# let getright e = match e with Right rt -> rt | Left _ -> failwith "not a Right";;
-# 100 + getright (let v = reset (fun p () -> Right (10 + shift p (fun k -> Left k))) in getleft v 1);;
--->
-
-Example 7:
-
- ; (+ 100 (reset (+ 10 (shift k (k (k 1)))))) ~~> 121
-
- app2 (op2 plus) (var hundred)
- (reset (app2 (op2 plus) (var ten)
- (shift (\k. app (var k) (app (var k) (var one))))))
-
- Continuation_monad.(let v = reset (
- let u = shift (fun k -> k 1 >>= fun x -> k x)
- in u >>= fun x -> unit (10 + x)
- ) in let w = v >>= fun x -> unit (100 + x)
- in run0 w)
- - : int = 121
-
-<!--
-
- print_endline "=== pa_monad's Continuation Tests ============";;
-
- (1, 5 = C.(run0 (unit 1 >>= fun x -> unit (x+4))) );;
- (2, 9 = C.(run0 (reset (unit 5 >>= fun x -> unit (x+4)))) );;
- (3, 9 = C.(run0 (reset (abort 5 >>= fun y -> unit (y+6)) >>= fun x -> unit (x+4))) );;
- (4, 9 = C.(run0 (reset (reset (abort 5 >>= fun y -> unit (y+6))) >>= fun x -> unit (x+4))) );;
- (5, 27 = C.(run0 (
- let c = reset(abort 5 >>= fun y -> unit (y+6))
- in reset(c >>= fun v1 -> abort 7 >>= fun v2 -> unit (v2+10) ) >>= fun x -> unit (x+20))) );;
-
- (7, 117 = C.(run0 (reset (shift (fun sk -> sk 3 >>= sk >>= fun v3 -> unit (v3+100) ) >>= fun v1 -> unit (v1+2)) >>= fun x -> unit (x+10))) );;
-
- (8, 115 = C.(run0 (reset (shift (fun sk -> sk 3 >>= fun v3 -> unit (v3+100)) >>= fun v1 -> unit (v1+2)) >>= fun x -> unit (x+10))) );;
-
- (12, ["a"] = C.(run0 (reset (shift (fun f -> f [] >>= fun t -> unit ("a"::t) ) >>= fun xv -> shift (fun _ -> unit xv)))) );;
-
-
- (0, 15 = C.(run0 (let f k = k 10 >>= fun v-> unit (v+100) in reset (callcc f >>= fun v -> unit (v+5)))) );;
-
--->
+You may encounter references to `prompt` and `control`. These are variants of `reset` and `shift` that differ in only subtle ways. As we said, there are lots of variants of these that we're not going to try to survey.
+
+We talked before about `abort`. This can be expressed in terms of `reset` and `shift`. At the end of our discussion of abort, we said that this diagram:
+
+ let foo x =
+ +---try begin----------------+
+ | (if x = 1 then 10 |
+ | else abort 20 |
+ | ) + 100 |
+ +---end----------------------+
+ in (foo 2) + 1000;;
+
+could be written in Scheme with either:
+
+ #lang racket
+ (require racket/control)
+
+ (let ([foo (lambda (x)
+ (reset
+ (+
+ (if (eqv? x 1) 10 (abort 20))
+ 100)))])
+ (+ (foo 2) 1000))
+
+or:
+
+ #lang racket
+ (require racket/control)
+
+ (let ([foo (lambda (x)
+ (reset
+ (+
+ (shift k
+ (if (eqv? x 1) (k 10) 20))
+ 100)))])
+ (+ (foo 2) 1000))
+
+That shows you how `abort` can be expressed in terms of `shift`. Rewriting the Scheme code into a more OCaml-ish syntax, it might look something like this:
+
+ let foo x = reset (shift k -> if x = 1 then k 10 else 20) + 100) in
+ foo 2 + 1000
+
+However, OCaml doesn't have any continuation operators in its standard deployment. If you [[installed Oleg's delimcc library|/rosetta3/#delimcc]], you can use the previous code after first doing this:
+
+ # #directory "+../delimcc";;
+ # #load "delimcc.cma";;
+ # let reset_label = ref None;;
+ # let reset body = let p = Delimcc.new_prompt () in
+ reset_label := Some p; let res = Delimcc.push_prompt p body in reset_label := None; res;;
+ # let shift fun_k = match !reset_label with
+ | None -> failwith "shift must be inside reset"
+ | Some p -> Delimcc.shift p fun_k;;
+
+Also, the previous code has to be massaged a bit to have the right syntax. What you really need to write is:
+
+ let foo x = reset (fun () -> shift (fun k -> if x = 1 then k 10 else 20) + 100) in
+ foo 2 + 1000
+
+That will return `1020` just like the Scheme code does. If you said `... foo 1 + 1000`, you'll instead get `1110`.
+
+That was all *delimited* continuation operators. There's also the **undelimited continuation operators**, which historically were developed first. Here you don't see the same kind of variety that you do with the delimited continuation operators. Essentially, there is just one full-strength undelimited continuation operator. But there are several different syntactic forms for working with it. (Also, a language might provide handicapped continuation operators alongside, or instead of, the full-strength one. Some loser languages don't even do that much.) The historically best-known of these is expressed in Scheme as `call-with-current-continuation`, or `call/cc` for short. But we think it's a bit easier to instead use the variant `let/cc`. The following code is equivalent, and shows how these two forms relate to each other:
+
+ (let/cc k ...)
+
+ (call/cc (lambda (k) ...))
+
+`(let/cc k ...)` is a lot like `(shift k ...)` (or in the OCaml version, `shift (fun k -> ...)`), except that it doesn't need a surrounding `reset ( ... )` (in OCaml, `reset (fun () -> ...)`). For the undelimited continuation operator, the box is understood to be *the whole rest of the top-level computation*. If you're running a file, that's all the rest of the file that would have been executed after the syntactic hole filled by `(let/cc k ...)`. With `(shift k ...)`, the code that gets bound to `k` doesn't get executed unless you specifically invoke `k`; but `let/cc` works differently in this respect. Thus:
+
+ (+ 100 (let/cc k 1))
+
+returns `101`, whereas:
+
+ (reset (+ 10 (shift k 1)))
+
+only returns `1`. It is possible to duplicate the behavior of `let/cc` using `reset`/`shift`, but you have to structure your code in certain ways to do it. In order to duplicate the behavior of `reset`/`shift` using `let/cc`, you need to also make use of a mutable reference cell. So in that sense delimited continuations are more powerful and undelimited continuations are sort-of a special case.
+
+(In the OCaml code above for using delimited continuations, there is a mutable reference cell `reset_label`, but this is just for convenience. Oleg's library is designed for use with _multiple_ reset blocks having different labels, then when you invoke `shift` you have to specify which labeled reset block you want to potentially skip the rest of. We haven't introduced that complexity into our discussion, so for convenience we worked around it in showing you how to use `reset` and `shift` in OCaml. And the mutable reference cell was only playing the role of enabling us to work around the need to explicitly specify the `reset` block's label.)
+
+Here are some examples of using these different continuation operators. The continuation that gets bound to `k` will be in bold. I'll use an OCaml-ish syntax because that's easiest to read, but these examples don't work as-is in OCaml. The `reset`/`shift` examples need to be massaged into the form displayed above for OCaml; and the `let/cc` examples don't work in OCaml because that's not provided. Alternatively, you could massage all of these into Scheme syntax. You shouldn't find that hard.
+
+1. <pre><b>100 + </b>let/cc k (10 + 1)</pre>
+ This evaluates to `111`. Nothing exotic happens here.
+
+2. <pre><b>100 + </b>let/cc k (10 + k 1)</pre>
+ This evaluates to `101`. See also example 11, below.
+
+3. <pre><b>let p = </b>let/cc k (1,k) <b>in
+ let y = snd p (2, ident) in
+ (fst p, y)</b></pre>
+ In the first line, we bind the continuation function (the bold code) to `k` and then bind the pair of `1` and that function to the variable `p`.
+ In the second line, we extract the continuation function from the pair `p` and apply it to the argument `(2, ident)`. That results in the following code being run:
+ <pre><b>let p = </b>(2, ident) <b>in
+ let y = snd p (2, ident) in
+ (fst p, y)</b></pre>
+ which in turn results in the nested pair `(2, (2, ident))`.
+ Notice how the first time through, when `p`'s second element is a continuation, applying it to an argument is a bit like time-travel? The metaphysically impossible kind of time-travel, where you can change what happened. The second time through, `p` gets bound to a different pair, whose second element is the ordinary `ident` function.
+
+4. <pre><b>1000 + (100 + </b>abort 11<b>)</b></pre>
+ Here the box is implicit, understood to be the rest of the code. The result is just the abort value `11`, because the bold code is skipped.
+
+5. <pre>1000 + reset <b>(100 + </b>abort 11<b>)</b></pre>
+ Here the box or delimiter is explicitly specified. The bold code is skipped, but the outside code `1000 + < >` is still executed, so we get `1011`.
+
+6. <pre>1000 + reset <b>(100 + </b>shift k (10 + 1)<b>)</b></pre>
+ Equivalent to preceding. We bind the bold code to `k` but then never apply `k`, so the value `10 + 1` is supplied directly to the outside code `1000 + < >`, resulting in `1011`.
+
+7. <pre>1000 + reset <b>(100 + </b>shift k (k (10 + 1))<b>)</b></pre>
+ Here we do invoke the captured continuation, so what gets passed to the outside code `1000 + < >` is `k (10 + 1)`, that is, `(100 + (10 + 1))`. Result is `1111`.
+ In general, if the last thing that happens inside a `shift k ( ... )` body is that `k` is applied to an argument, then we do continue running the bold code between `shift k ( ... )` and the edge of the `reset` box.
+
+8. <pre>1000 + reset <b>(100 + </b>shift k (10 + k 1)<b>)</b></pre>
+ This also results in `1111`, but via a different path than the preceding. First, note that `k` is bound to `100 + < >`. So `k 1` is `101`. Then `10 + k 1` is `10 + 101`. Then we exit the body of `shift k ( ... )`, without invoking `k` again, so we don't add `100` any more times. Thus we pass `10 + 101` to the outside code `1000 + < >`. So the result is `1000 + (10 + 101)` or `1111`. (Whereas in the preceding example, the result was `1000 + (100 + 11)`. The order in which the operations are performed is different. If we used a non-commutative operation instead of `+`, the results of these two examples would be different from each other.)
+
+9. <pre>1000 + reset <b>(100 + </b>shift k (k)<b>)</b> 1</pre>
+ Here `k` is bound to `100 + < >`. That function `k` is what's returned by the `shift k ( ... )` block, and since `k` isn't invoked (applied) when doing so, the rest of the bold `reset` block is skipped (for now). So we resume the outside code `1000 + < > 1`, with what fills the gap `< >` being the function that was bound to `k`. Thus this is equivalent to `1000 + (fun x -> 100 + x) 1` or `1000 + 101` or `1101`.
+
+10. <pre>1000 + reset <b>(100 + </b>shift k (k (k 1))<b>)</b></pre>
+ Here `k` is bound to `100 + < >`. Thus `k 1` is `101`. Now there are two ways to think about what happens next. (Both are valid.) One way to think is that since the `shift` block ends with an additional outermost application of `k`, then as described in example 7 above, we continue through the bold code with the value `k 1` or `101`. Thus we get `100 + 101`, and then we continue with the outermost code `1000 + < >`, getting `1000 + (100 + 101)`, or `1201`. The other way to think is that since `k` is `100 + < >`, and `k 1` is `101`, then `k (k 1)` is `201`. Now we leave the `shift` block *without* executing the bold code a third time (we've already taken account of the two applications of `k`), resuming with the outside code `1000 + < >`, thereby getting `1000 + 201` as before.
+
+11. Here's a comparison of `let/cc` to `shift`. Recall example 2 above was:
+ <pre><b>100 + </b>let/cc k (10 + k 1)</pre>
+ which evaluated to `101`. The parallel code where we instead capture the continuation using `shift k ( ... )` would look like this:
+ <pre>reset <b>(100 + </b>shift k (10 + k 1)<b>)</b></pre>
+ But this evaluates differently. In the `let/cc` example, `k` is bound to the rest of the computation *including its termination*, so after executing `k 1` we never come back and finish with `10 + < >`. A `let/cc`-bound `k` never returns to the context where it was invoked. Whereas the `shift`-bound `k` only includes up to the edge of the `reset` box --- here, the rest of the computation, but *not including* its termination. So after `k 1`, if there is still code inside the body of `shift`, as there is here, we continue executing it. Thus the `shift` code evaluates to `111` not to `101`.
+
+ Thus code using `let/cc` can't be *straightforwardly* translated into code using `shift`. It can be translated, but the algorithm will be more complex.