XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=cps_and_continuation_operators.mdwn;h=72b0034235e0c2be0e891c350ed168dee3b6360f;hp=314fb4341e6142aa1d7c2914d827f7a383e6112c;hb=399197e0af7c4b80f326be60ff2c0caae74b3687;hpb=6f77f31cdd1687494ce1f52277e5486ff4293540
diff git a/cps_and_continuation_operators.mdwn b/cps_and_continuation_operators.mdwn
index 314fb434..72b00342 100644
 a/cps_and_continuation_operators.mdwn
+++ b/cps_and_continuation_operators.mdwn
@@ 3,7 +3,8 @@
CPS Transforms
==============
We've already approached some tasks now by programming in **continuationpassing style.** We first did that with tuples at the start of term, and then with the v5 lists in [[week4]], and now more recently and selfconsciously when discussing [aborts](/couroutines_and_aborts), and [the "abSd" task](from_list_zippers_to_continuations/). and the use of `tree_monadize` specialized to the Continuation monad, which required us to supply an initial continuation.
+We've already approached some tasks now by programming in **continuationpassing style.** We first did that with tuples at the start of term, and then with the v5 lists in [[week4]], and now more recently and selfconsciously when discussing [aborts](/couroutines_and_aborts),
+and [the "abSd" task](/from_list_zippers_to_continuations). and the use of `tree_monadize` specialized to the Continuation monad, which required us to supply an initial continuation.
In our discussion of aborts, we showed how to rearrange code like this:
@@ 86,19 +87,19 @@ And here is another:
These transforms have some interesting properties. One is thatassuming we never reduce inside a lambda term, but only when redexes are present in the outermost levelthe formulas generated by these transforms will always only have a single candidate redex to be reduced at any stage. In other words, the generated expressions dictate in what order the components from the original expressions will be evaluated. As it happens, the first transform above forces a *callbyname* reduction order: assuming `M N` to be a redex, redexes inside `N` will be evaluated only after `N` has been substituted into `M`. And the second transform forces a *callbyvalue* reduction order. These reduction orders will be forced no matter what the native reduction order of the interpreter is, just so long as we're only allowed to reduce redexes not underneath lambdas.
Plotkin did important early work with CPS transforms (around 1975), and they are now a staple of academic computer science.
+Plotkin did important early work with CPS transforms, and they are now a staple of academic computer science. (See the end of his 1975 paper [Callbyname, callbyvalue, and the lambdacalculus](http://homepages.inf.ed.ac.uk/gdp/publications/cbn_cbv_lambda.pdf).)
Here's another interesting fact about these transforms. Compare the translations for variables and applications in the callbyvalue transform:
[x] > \k. k x
[M N] > \k. [M] (\m. [N] (\n. m n k))
To the implementations we proposed for `unit` and `bind` when developing a Continuation monads, for example [here](/list_monad_as_continuation_monad). I'll relabel some of the variable names to help the comparison:
+to the implementations we proposed for `unit` and `bind` when developing a Continuation monads, for example [here](/list_monad_as_continuation_monad). I'll relabel some of the variable names to help the comparison:
let cont_unit x = fun k > k x
 let cont_bind M N = fun k > M (fun m > N m k)
+ let cont_bind N M = fun k > N (fun n > M n k)
The transform for `x` is just `cont_unit x`! And the transform for `M N` is, though not here exactly the same as `cont_bind M N`, quite reminiscent of it. (I don't yet know whether there's an easy and satisfying explanation of why these two diverge as they do.)
+The transform for `x` is just `cont_unit x`! And the transform for `M N` is, though not here exactly the same as `cont_bind N M`, quite reminiscent of it. (I don't yet know whether there's an easy and satisfying explanation of why these two are related as they are.)
Doing CPS transforms by hand is very cumbersome. (Try it.) But you can leverage our lambda evaluator to help you out. Here's how to do it. From here on out, we'll be working with and extending the callbyvalue CPS transform set out above:
@@ 124,15 +125,16 @@ Or, using the lambda evaluator, that is:
~~> \k. k (succ x)
Some other handy tools:
 let app3 = \a b c. app (app a b) c in
 let app4 = \a b c d. app (app (app a b) c) d in
+
+ let app2 = \a b c. app (app a b) c in
+ let app3 = \a b c d. app (app (app a b) c) d in
let op2 = \op. \u. u (\a v. v (\b k. k (op a b))) in
let op3 = \op. \u. u (\a v. v (\b w. w (\c k. k (op a b c)))) in
...
Then, for instance, [plus x y] would be rendered in the lambda evaluator as:
 app3 (op2 plus) (var x) (var y)
+ app2 (op2 plus) (var x) (var y)
~~> \k. k (plus x y)
To finish off a CPS computation, you have to supply it with an "initial" or "outermost" continuation. (This is somewhat like "running" a monadic computation.) Usually you'll give the identity function, representing that nothing further happens to the continuationexpecting value.
@@ 148,7 +150,7 @@ I won't give the CPS transform for `callcc` itself, but instead for the complex
[callcc (\k. body)] = \outk. (\k. [body] outk) (\v localk. outk v)
The behavior of `callcc` is this. The whole expression `callcc (\k. body)`, call it C, is being evaluated in a context, call it E[\_]. When we convert to CPS form, the continuation of this occurrence of C will be bound to the variable `outk`. What happens then is that we bind the expression `\v localk. outk v` to the variable `k` and evaluate [body], passing through to it the existing continuation `outk`. Now if `body` is just, for example, `x`, then its CPS transform [x] will be `\j. j x` and this will accept the continuation `outk` and feed it `x`, and we'll continue on with nothing unusual occurring. If on the other hand `body` makes use of the variable `k`, what happens then? For example, suppose `body` includes `foo (k v)`. In the reduction of the CPS transform `[foo (k v)]`, `v` will be passed to `k` which as we said is now `\v localk. outk v`. The continuation of that applicationwhat happens to `k v` after it's evaluated and `foo` gets access to itwill be bound next to `localk`. But notice that this `localk` is discarded. The computation goes on without it. Instead, it just continues evaluating `outk v`, where as we said `outk` is the outside continuation E[\_] of the whole `callcc (\k. body)` invocation.
+The behavior of `callcc` is this. The whole expression `callcc (\k. body)`, call it C, is being evaluated in a context, call it E[\_]. When we convert to CPS form, the continuation of this occurrence of C will be bound to the variable `outk`. What happens then is that we bind the expression `\v localk. outk v` to the variable `k` and evaluate [body], passing through to it the existing continuation `outk`. Now if `body` is just, for example, `x`, then its CPS transform [x] will be `\j. j x` and this will accept the continuation `outk` and feed it `x`, and we'll continue on with nothing unusual occurring. If on the other hand `body` makes use of the variable `k`, what happens then? For example, suppose `body` includes `foo (k v)`. In the reduction of the CPS transform `[foo (k v)]`, `v` will be passed to `k` which as we said is now `\v localk. outk v`. The continuation of that applicationwhat is scheduled to happen to `k v` after it's evaluated and `foo` gets access to itwill be bound next to `localk`. But notice that this `localk` is discarded. The computation goes on without it. Instead, it just continues evaluating `outk v`, where as we said `outk` is the outside continuation E[\_] of the whole `callcc (\k. body)` invocation.
So in other words, since the continuation in which `foo` was to be applied to the value of `k v` was discarded, that application never gets evaluated. We escape from that whole block of code.
@@ 156,7 +158,7 @@ It's important to understand that `callcc` binds `k` to a pipe into the continua
So too will examples. We'll give some examples, and show you how to try them out in a variety of formats:
(i) using the lambda evaluator to check how the CPS transforms reduce
+1. using the lambda evaluator to check how the CPS transforms reduce
To do this, you can use the following helper function:
@@ 165,39 +167,39 @@ So too will examples. We'll give some examples, and show you how to try them out
Used like this: [callcc (\k. body)] = `callcc (\k. BODY)`, where `BODY` is [body].
(ii) using a `callcc` operation on our Continuation monad
+2. using a `callcc` operation on our Continuation monad
This is implemented like this:
let callcc body = fun outk > body (fun v localk > outk v) outk
 GOTCHAS??
+
(iii) `callcc` was originally introduced in Scheme. There it's written `call/cc` and is an abbreviation of `callwithcurrentcontinuation`. Instead of the somewhat bulky form:
+3. `callcc` was originally introduced in Scheme. There it's written `call/cc` and is an abbreviation of `callwithcurrentcontinuation`. Instead of the somewhat bulky form:
 (call/cc (lambda (k) ...))
+ (call/cc (lambda (k) ...))
I prefer instead to use the lighter, and equivalent, shorthand:
+ I prefer instead to use the lighter, and equivalent, shorthand:
 (let/cc k ...)
+ (let/cc k ...)
Callcc examples

+Callcc/letcc examples
+
First, here are two examples in Scheme:
(+ 100 (let/cc k (+ 10 1)))

This binds the continuation `outk` of the underlined expression to `k`, the computes `(+ 10 1) and delivers that to `outk` in the normal way (not through `k`). No unusual behavior. It evaluates to `111.
+This binds the continuation `outk` of the underlined expression to `k`, then computes `(+ 10 1)` and delivers that to `outk` in the normal way (not through `k`). No unusual behavior. It evaluates to `111`.
Now if we do instead:
+What if we do instead:
(+ 100 (let/cc k (+ 10 (k 1))))

Now, during the evaluation of `(+ 10 (k 1))`, we supply `1` to `k`. So then the local continuation, which delivers the value up to `(+ 10 _)` and so on, is discarded. Instead `1` gets supplied to the outer continuation in place when `let/cc` was invoked. That will be `(+ 100 _)`. When `(+ 100 1)` is evaluated, there's no more of the computation left to evaluate. So the answer here is `101`.
+This time, during the evaluation of `(+ 10 (k 1))`, we supply `1` to `k`. So then the local continuation, which delivers the value up to `(+ 10 [_])` and so on, is discarded. Instead `1` gets supplied to the outer continuation in place when `let/cc` was invoked. That will be `(+ 100 [_])`. When `(+ 100 1)` is evaluated, there's no more of the computation left to evaluate. So the answer here is `101`.
You are not restricted to calling a bound continuation only once, nor are you restricted to calling it only inside of the `call/cc` (or `let/cc`) block. For example, you can do this:
@@ 205,7 +207,7 @@ You are not restricted to calling a bound continuation only once, nor are you re
(cons (car p) ((cdr p) (cons 2 (lambda (x) x)))))
; evaluates to '(2 2 . #)
What happens here? First, we capture the continuation where `p` is about to be assigned a value. Inside the `let/cc` block, we create a pair consisting of `1` and the captured continuation. This pair is bound to p. We then proceed to extract the components of the pair. The head (`car`) goes into the start of a tuple we're building up. To get the next piece of the tuple, we extract the second component of `p` (this is the bound continuation `k`) and we apply it to a pair consisting of `2` and the identity function. Supplying arguments to `k` takes us back to the point where `p` is about to be assigned a value. The tuple we had formerly been building, starting with `1`, will no longer be accessible because we didn't bring along with us any way to refer to it, and we'll never get back to the context where we supplied an argument to `k`. Now `p` gets assigned not the result of `(let/cc k (cons 1 k))` again, but instead, the new pair that we provided: `'(2 . #)`. Again we proceed to build up a tuple: we take the first element `2`, then we take the second element (now the identity function), and feed it a pair `'(2 . #)`, and since it's an argument to the identity procedure that's also the result. So our final result is a nest pair, whose first element is `2` and whose second element is the pair `'(2 . #)`. Racket displays this nested pair like this:
+What happens here? First, we capture the continuation where `p` is about to be assigned a value. Inside the `let/cc` block, we create a pair consisting of `1` and the captured continuation. This pair is bound to p. We then proceed to extract the components of the pair. The head (`car`) goes into the start of a tuple we're building up. To get the next piece of the tuple, we extract the second component of `p` (this is the bound continuation `k`) and we apply it to a pair consisting of `2` and the identity function. Supplying arguments to `k` takes us back to the point where `p` is about to be assigned a value. The tuple we had formerly been building, starting with `1`, will no longer be accessible because we didn't bring along with us any way to refer to it, and we'll never get back to the context where we supplied an argument to `k`. Now `p` gets assigned not the result of `(let/cc k (cons 1 k))` again, but instead, the new pair that we provided: `'(2 . #)`. Again we proceed to build up a tuple: we take the first element `2`, then we take the second element (now the identity function), and feed it a pair `'(2 . #)`, and since it's an argument to the identity procedure that's also the result. So our final result is a nested pair, whose first element is `2` and whose second element is the pair `'(2 . #)`. Racket displays this nested pair like this:
'(2 2 . #)
@@ 216,21 +218,21 @@ In the lambda evaluator:
let var = \x (\k. k x) in
let lam = \x_body (\k. k (\x. x_body x)) in
let app = \m n. (\k. m (\m. n (\n. m n k))) in
 let app3 = \a b c. app (app a b) c in
 let app4 = \a b c d. app (app (app a b) c) d in
+ let app2 = \a b c. app (app a b) c in
+ let app3 = \a b c d. app (app (app a b) c) d in
let op1 = \op. \u. u (\a k. k (op a)) in
let op2 = \op. \u. u (\a v. v (\b k. k (op a b))) in
let op3 = \op. \u. u (\a v. v (\b w. w (\c k. k (op a b c)))) in
let callcc = \k_body. \outk. (\k. (k_body k) outk) (\v localk. outk v) in
; (+ 100 (let/cc k (+ 10 1))) ~~> 111
 app3 (op2 plus) (var hundred) (callcc (\k. app3 (op2 plus) (var ten) (var one)))
+ app2 (op2 plus) (var hundred) (callcc (\k. app2 (op2 plus) (var ten) (var one)))
; evaluates to \k. k (plus hundred (plus ten one))
Next:
; (+ 100 (let/cc k (+ 10 (k 1)))) ~~> 101
 app3 (op2 plus) (var hundred) (callcc (\k. app3 (op2 plus) (var ten) (app (var k) (var one))))
+ app2 (op2 plus) (var hundred) (callcc (\k. app2 (op2 plus) (var ten) (app (var k) (var one))))
; evaluates to \k. k (plus hundred one)
We won't try to do the third example in this framework.
@@ 244,7 +246,7 @@ Now what we want to do is something like this:
# C.(run0 (100 + callcc (fun k > 10 + 1)));;
`run0` is a special function in the Continuation monad that runs a value of that monad using the identity function as its initial continuation. The above expression won't typeecheck, for several reasons. First, we're trying to add 100 to `callcc (...)` but the latter is a `Continuation.m` value, not an `int`. So we have to do this instead:
+`run0` is a special function in the Continuation monad that runs a value of that monad using the identity function as its initial continuation. The above expression won't typecheck, for several reasons. First, we're trying to add 100 to `callcc (...)` but the latter is a `Continuation.m` value, not an `int`. So we have to do this instead:
# C.(run0 (callcc (fun k > 10 + 1) >>= fun i > 100 + i));;
@@ 266,22 +268,144 @@ That won't work because `k 1` doesn't have type `int`, but we're trying to add i
This also works and as you can see, delivers the expected answer `101`.
At the moment, I'm not able to get the third example working with the monadic library. I thought that this should do it, but it doesn't typecheck:
+The third example is more difficult to make work with the monadic library, because its types are tricky. I was able to get this to work, which uses OCaml's "polymorphic variants." These are generally more relaxed about typing. There may be a version that works with regular OCaml types, but I haven't yet been able to identify it. Here's what does work:
+
+ # C.(run0 (callcc (fun k > unit (1,`Box k)) >>= fun (p1,`Box p2) > p2 (2,`Box unit) >>= fun p2' > unit (p1,p2')));;
+  : int * (int * [ `Box of 'b > ('a, 'b) C.m ] as 'b) as 'a =
+ (2, (2, `Box ))
+
+
+
+Some callcc/letcc exercises
+
+
+Here are a series of examples from *The Seasoned Schemer*, which we recommended at the start of term. It's not necessary to have the book to follow the exercises, though if you do have it, its walkthroughs will give you useful assistance.
 # C.(run0 (callcc (fun k > unit (1,k)) >>= fun (p1,p2) > p2 (2,unit) >>= fun p2' > unit (p1,p2')));;
+For reminders about Scheme syntax, see [here](/assignment8/) and [here](/week1/) and [here](/translating_between_ocaml_scheme_and_haskell). Other resources are on our [[Learning Scheme]] page.
If we figure this out later (or anyone else does), we'll come back and report.
+Most of the examples assume the following preface:
+ #lang racket
+
+ (define (atom? x)
+ (and (not (pair? x)) (not (null? x))))
+
+Now try to figure out what this function does:
+
+ (define alpha
+ (lambda (a lst)
+ (let/cc k ; now what will happen when k is called?
+ (letrec ([aux (lambda (l)
+ (cond
+ [(null? l) '()]
+ [(eq? (car l) a) (k (aux (cdr l)))]
+ [else (cons (car l) (aux (cdr l)))]))])
+ (aux lst)))))
+
+Here is [the answer](/hints/cps_hint_1), but try to figure it out for yourself.
+
+Next, try to figure out what this function does:
+
+ (define beta
+ (lambda (lst)
+ (let/cc k ; now what will happen when k is called?
+ (letrec ([aux (lambda (l)
+ (cond
+ [(null? l) '()]
+ [(atom? (car l)) (k (car l))]
+ [else (begin
+ ; what will the value of the next line be? why is it ignored?
+ (aux (car l))
+ (aux (cdr l)))]))])
+ (aux lst)))))
+
+Here is [the answer](/hints/cps_hint_2), but try to figure it out for yourself.
+
+Next, try to figure out what this function does:
+
+ (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 nonatom?
+ [(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
===========================
`callcc` is what's known as an *undelimited control operator*. That is, the continuations `outk` that get bound inside our `k`s behave as though they include all the code from the `call/cc ...` out to *and including* the end of the program.
+Here again is the CPS transform for `callcc`:
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 the latter have been formulated.
+ [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 workaround. Also. if you've got a copy of *The Seasoned Schemer*, see the comparison of let/cc vs. "collectorusing" (that is, partly CPS) functions at pp. 155164.)
The most wellbehaved 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.
+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 wellbehaved 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:
@@ 306,7 +430,7 @@ I set (2) aside a moment ago. The story we just told is a bit too simple because
For instance, in Scheme this:
(require racket/control)

+
(reset
(let ([x 1])
(+ 10 (shift k x))))
@@ 347,22 +471,22 @@ Using `shift` and `reset` operators in OCaml, this would look like this:
let reset = Delimcc.push_prompt p;;
let shift = Delimcc.shift p;;
let abort = Delimcc.abort p;;

+
let foo x =
reset(fun () >
 shift(fun continue >
 if x = 1 then continue 10
 else 20
 ) + 100
+ shift(fun continue >
+ if x = 1 then continue 10
+ else 20
+ ) + 100
)
in foo 2 + 1000;;
 : int = 1020
If instead at the end we did `in foo 1 + 1000`, we'd get the result `1110`.
+If instead at the end we did `... foo 1 + 1000`, we'd get the result `1110`.
The above OCaml code won't work out of the box; you have to compile and install a special library that Oleg wrote. We discuss it on our [translation page](/translating_between_ocaml_scheme_and_haskell). If you can't get it working, then you can play around with `shift` and `reset` in Scheme instead. Or in the Continuation monad. Or using CPS transforms of your code, with the help of the lambda evaluator.
The relevant CPS transforms will be performed by these helper functions:
+You can make the lambda evaluator perform the required CPS transforms with these helper functions:
let reset = \body. \outk. outk (body (\i i)) in
let shift = \k_body. \midk. (\k. (k_body k) (\i i)) (\a localk. localk (midk a)) in
@@ 371,14 +495,14 @@ The relevant CPS transforms will be performed by these helper functions:
You use these like so:
* [prompt m] is `prompt M` where `M` is [m]
* [shift k M] is `shift (\k. M)` where `M` is [m]
* and [abort M] is `abort M` where `M` is [m]
+* [reset body] is `reset BODY` where `BODY` is [body]
+* [shift k body] is `shift (\k. BODY)` where `BODY` is [body]
+* and [abort value] is `abort VALUE` where `VALUE` is [value]
There are also `reset` and `shift` and `abort` operations in the Continuation monad in our OCaml [[monad library]]. You can check the code for details.
As we said, there are many varieties of delimited continuations. Another common pair is `prompt` and `control`. There's no difference in meaning between `prompt` and `reset`; it's just that people tend to say `reset` when talking about `shift` and `prompt` when talking about `control`. `control` acts subtly differently from `shift`. In the uses you're likely to make as you're just learning about continuations, you won't see any difference. If you'll do more research in this vicinity, you'll soon enough learn about the differences.
+As we said, there are many varieties of delimited continuations. Another common pair is `prompt` and `control`. There's no difference in meaning between `prompt` and `reset`; it's just that people tend to say `reset` when talking about `shift`, and `prompt` when talking about `control`. `control` acts subtly differently from `shift`. In the uses you're likely to make as you're just learning about continuations, you won't see any difference. If you'll do more research in this vicinity, you'll soon enough learn about the differences.
(You can start by reading [the Racket docs](http://docs.racketlang.org/reference/cont.html?q=shift&q=do#%28part._.Classical_.Control_.Operators%29).)
@@ 394,146 +518,137 @@ In collecting these CPS transforms and implementing the monadic versions, we've
Examples of shift/reset/abort

Here are some more examples of using delimited control operators. We present first a Scheme formulation; then we compute the same result using CPS and the lambda evaluator.
+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.)
 ; (+ 100 (+ 10 (abort 1))) ~~> 1
 app3 (op2 plus) (var hundred)
 (app3 (op2 plus) (var ten) (abort (var one)))
+Example 1:
 ; (+ 100 (prompt (+ 10 (abort 1)))) ~~> 101
 app3 (op2 plus) (var hundred)
 (prompt (app3 (op2 plus) (var ten) (abort (var one))))
+ ; (+ 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
 ; (+ 1000 (prompt (+ 100 (shift k (+ 10 1))))) ~~> 1011
 app3 (op2 plus) (var thousand)
 (prompt (app3 (op2 plus) (var hundred)
 (shift (\k. ((op2 plus) (var ten) (var one))))))
+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:
 ; (+ 1000 (prompt (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111
 app3 (op2 plus) (var thousand)
 (prompt (app3 (op2 plus) (var hundred)
 (shift (\k. (app (var k) ((op2 plus) (var ten) (var one)))))))
+ # Continuation_monad.(run0(
+ reset (
+ abort 11 >>= fun i >
+ unit (100+i) >>= fun j >
+ unit (1000+j))));;
+  : int = 11
 ; (+ 1000 (prompt (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently
 app3 (op2 plus) (var thousand)
 (prompt (app3 (op2 plus) (var hundred)
 (shift (\k. ((op2 plus) (var ten) (app (var k) (var one)))))))
 ; (+ 100 ((prompt (+ 10 (shift k k))) 1)) ~~> 111
 app3 (op2 plus) (var hundred)
 (app (prompt (app3 (op2 plus) (var ten)
 (shift (\k. (var k))))) (var one))
+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
 ; (+ 100 (prompt (+ 10 (shift k (k (k 1)))))) ~~> 121
 app3 (op2 plus) (var hundred)
 (prompt (app3 (op2 plus) (var ten)
 (shift (\k. app (var k) (app (var k) (var one))))))
+Example 3:
+ ; (+ 1000 (reset (+ 100 (shift k (+ 10 1))))) ~~> 1011
More:
+ app2 (op2 plus) (var thousand)
+ (reset (app2 (op2 plus) (var hundred)
+ (shift (\k. ((op2 plus) (var ten) (var one))))))
 (* (+ 1000 (prompt (+ 100 (shift k (+ 10 1))))) ~~> 1011 *)
 let example1 () : int =
 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 run w)
+ 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
 (* (+ 1000 (prompt (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111 *)
 let example2 () =
 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 run w)
+Example 4:
 (* (+ 1000 (prompt (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently *)
 let example3 () =
 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 run w)
+ ; (+ 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]
 (* (+ 100 ((prompt (+ 10 (shift k k))) 1)) ~~> 111 *)
 (* not sure if this example can be typed without a sumtype *)
+To demonstrate the different adding order between Examples 4 and 5, we use `::` in the OCaml version instead of `+`. Here is Example 5:
 (* (+ 100 (prompt (+ 10 (shift k (k (k 1)))))) ~~> 121 *)
 let example5 () : int =
 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 run w)
+ ; (+ 1000 (reset (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently
More:
+ 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]
 module C = Continuation_monad;;
+Example 6:
 print_endline "=== test TreeT(Continuation).distribute ==================";;
+ ; (+ 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 asis 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
 let id : 'z. 'z > 'z = fun x > x
+
 let example n : (int * int) =
 Continuation_monad.(let u = callcc (fun k >
 (if n < 0 then k 0 else unit [n + 100])
 (* all of the following is skipped by k 0; the end type int is k's input type *)
 >>= fun [x] > unit (x + 1)
 )
 (* k 0 starts again here, outside the callcc (...); the end type int * int is k's output type *)
 >>= fun x > unit (x, 0)
 in run0 u)


 (* (+ 1000 (prompt (+ 100 (shift k (+ 10 1))))) ~~> 1011 *)
 let example1 () : int =
 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)

 (* (+ 1000 (prompt (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111 *)
 let example2 () =
 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)

 (* (+ 1000 (prompt (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently *)
 let example3 () =
 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)

 (* (+ 100 ((prompt (+ 10 (shift k k))) 1)) ~~> 111 *)
 (* not sure if this example can be typed without a sumtype *)

 (* (+ 100 (prompt (+ 10 (shift k (k (k 1)))))) ~~> 121 *)
 let example5 () : int =
 Continuation_monad.(let v = reset (
 let u = shift (fun k > k 1 >>= k)
 in u >>= fun x > unit (10 + x)
 ) in let w = v >>= fun x > unit (100 + x)
 in run0 w)

 ;;

More:

 print_endline "=== test bare Continuation ============";;

 (1011, 1111, 1111, 121);;
 (example1(), example2(), example3(), example5());;
 ((111,0), (0,0));;
 (example ~+10, example ~10);;
+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
+