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7 Last week we saw how to turn a list zipper into a continuation-based
8 list processor. The function computed something we called "the task",
9 which was a simplified langauge involving control operators.
11 abSdS ~~> ababdS ~~> ababdababd
13 The task is to process the list from left to right, and at each "S",
14 double the list so far. Here, "S" is a kind of control operator, and
15 captures the entire previous computation. We also considered a
16 variant in which '#' delimited the portion of the list to be copied:
20 In this variant, "S" and "#" correspond to `shift` and `reset`, which
21 provide access to delimited continuations.
23 The expository logic of starting with this simplified task is the
24 notion that as lists are to trees, so is this task to full-blown
25 continuations. So to the extent that, say, list zippers are easier to
26 grasp than tree zippers, the task is easier to grasp than full
29 We then presented CPS transforms, and demonstrated how they provide
30 an order-independent analysis of order of evaluation.
32 In order to continue to explore continuations, we will proceed in the
33 following fashion: we introduce the traditional continuation monad,
34 and show how it solves the task, then generalize the task to
35 include doubling of both the left and the right context.
37 ## The continuation monad
39 In order to build a monad, we start with a Kleisli arrow.
41 Continuation monad: types: given some ρ, Mα => (α -> ρ) -> ρ
43 bind == \ufk. u(\x.fxk)
45 We'll first show that this monad solves the task, then we'll consider
46 the monad in more detail.
48 The unmonadized computation (without the shifty "S" operator) is
50 t1 = + a (+ b (+ c d)) ~~> abcd
52 where "+" is string concatenation and the symbol a is shorthand for
55 In order to use the continuation monad to solve the list task,
56 we choose α = ρ = [Char]. So "abcd" is a list of characters, and
57 a boxed list has type M[Char] == ([Char] -> [Char]) -> [Char].
59 Writing ¢ in between its arguments, t1 corresponds to the following
62 mt1 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (⇧+ ¢ ⇧c ¢ ⇧d))
64 We have to lift each functor (+) and each object (e.g., "b") into the
65 monad using mid (`⇧`), then combine them using monadic function
68 ¢ M N = \k -> M (\f -> N (\a -> k(f x)))
70 for the continuation monad.
72 The way in which we extract a value from a continuation box is by
73 applying it to a continuation; often, it is convenient to supply the
74 trivial continuation, the identity function \k.k = I. So in fact,
78 That is, the original computation is the monadic version applied to
79 the trivial continuation.
81 We can now add a shifty operator. We would like to replace just the
82 one element, and we will do just that in a moment; but in order to
83 simulate the original task, we'll have to take a different strategy
84 initially. We'll start by imagining a shift operator that combined
85 direction with the tail of the list, like this:
87 mt2 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ ⇧d))
89 We can now define a shift operator to perform the work of "S":
91 shift u k = u(\s.k(ks))
93 Shift takes two arguments: a string continuation u of type M[Char],
94 and a string continuation k of type [Char] -> [Char]. Since u is the
95 the argument to shift, it represents the tail of the list after the
96 shift operator. Then k is the continuation of the expression headed
97 by `shift`. So in order to execute the task, shift needs to invoke k
98 twice. The expression `\s.k(ks)` is just the composition of k with itself.
104 Let's just make sure that we have the left-to-right evaluation we were
105 hoping for by evaluating "abSdeSf":
107 mt3 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ (⇧+ ¢ ⇧d ¢ (⇧+ ¢ ⇧e ¢ (shift ⇧f)))))
111 mt3 I = "ababdeababdef" -- structure: (ababde)(ababde)f
116 For a reset operator #, we can have
118 # u k = k(u(\k.k)) -- ex.: ab#deSf ~~> abdedef
120 The reset operator executes the remainder of the list separately, by
121 giving it the trivial continuation (\k.k), then feeds the result to
122 the continuation corresponding to the position of the reset.
124 So the continuation monad solves the list task using continuations in
125 a way that conforms to our by-now familiar strategy of lifting a
126 computation into a monad, and then writing a few key functions (shift,
127 reset) that exploit the power of the monad.
129 ## Generalizing to the tree doubling task
131 Now we should consider what happens when we write a shift operator
132 that takes the place of a single letter.
134 mt2 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ ⇧d))
135 mt4 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (⇧+ ¢ shift' ¢ ⇧d))
137 Instead of mt2 (copied from above), we have mt4. So now the type of a
138 leaf (a boxed string, type M[Char]) is the same as the type of the new
139 shift operator, shift'.
143 This shift operator takes a continuation k of type [Char]->[Char], and
144 invokes it twice. Since k requires an argument of type [Char], we
145 need to use the first invocation of k to construction a [Char]; we do
146 this by feeding it a string. Since the task does not replace the
147 shift operator with any marker, we give the empty string "" as the
150 But now the new shift operator captures more than just the preceeding
151 part of the construction---it captures the entire context, including
152 the portion of the sequence that follows it. That is,
156 We have replaced "S" in "abSd" with "ab_d", where the underbar will be
157 replaced with the empty string supplied in the definition of shift'.
158 Crucially, not only is the prefix "ab" duplicated, so is the suffix
161 Things get interesting when we have more than one operator in the
162 initial list. What should we expect if we start with "aScSe"?
163 If we assume that when we evaluate each S, all the other S's become
164 temporarily inert, we expect a reduction path like
168 But note that the output has just as many S's as the input--if that is
169 what our reduction strategy delivers, then any initial string with
170 more than one S will never reach a normal form.
172 But that's not what the continuation operator shift' delivers.
174 mt5 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ shift' ¢ (⇧+ ¢ ⇧c ¢ (⇧+ ¢ shift' ¢ "e")))
176 mt5 I = "aacaceecaacaceecee" -- structure: "aacaceecaacaceecee"
180 This is considerably harder to understand than the original list task.
181 The key is figuring out in each case what function the argument k to
182 the shift operator gets bound to.
184 Let's go back to a simple one-shift example, "aSc". Let's trace what
185 the shift' operator sees as its argument k by replacing ⇧ and ¢ with
189 ⇧+ ¢ ⇧a ¢ (⇧+ ¢ shift' ¢ ⇧c) I
190 = \k.⇧+(\f.⇧a(\x.k(fx))) ¢ (⇧+ ¢ shift' ¢ ⇧c) I
191 = \k.(\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(\x.k(fx))) I
192 ~~> (\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(\x.I(fx)))
193 ~~> (\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(f))
194 ~~> ⇧+(\f.⇧a(\x.(\f.(⇧+ ¢ shift' ¢ ⇧c)(f))(fx))))
195 ~~> ⇧+(\f.⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(fx)))
196 = (\k.k+)(\f.⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(fx)))
197 ~~> ⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(+x))
198 = (\k.ka)(\x.(⇧+ ¢ shift' ¢ ⇧c)(+x))
199 ~~> (⇧+ ¢ shift' ¢ ⇧c)(+a)
200 = (\k.⇧+(\f.shift(\x.k(fx)))) ¢ ⇧c (+a)
201 = (\k.(\k.⇧+(\f.shift(\x.k(fx))))(\f.⇧c(\x.k(fx))))(+a)
202 ~~> (\k.⇧+(\f.shift(\x.k(fx))))(\f'.⇧c(\x'.(+a)(f'x')))
203 ~~> ⇧+(\f.shift(\x.(\f'.⇧c(\x'.(+a)(f'x')))(fx)))
204 ~~> ⇧+(\f.shift(\x.⇧c(\x'.(+a)((fx)x'))))
205 = (\k.k+)(\f.shift(\x.⇧c(\x'.(+a)((fx)x'))))
206 ~~> shift(\x.⇧c(\x'.(+a)((+x)x'))))
207 = shift(\x.(\k.kc)(\x'.(+a)((+x)x'))))
208 ~~> shift(\x.(+a)((+x)c))
211 So now we see what the argument of shift will be: a function k from
212 strings x to the string asc. So shift k will be k(k "") = aacc.
214 Ok, this is ridiculous. We need a way to get ahead of this deluge of
215 lambda conversion. We'll see how to understand what is going on
216 when we talk about quantifier raising in the next lecture.
218 ## Viewing Montague's PTQ as CPS
220 Montague's conception of determiner phrases as generalized quantifiers
221 is a limited form of continuation-passing. (See, e.g., chapter 4 of
222 Barker and Shan 2014.) Start by assuming that ordinary DPs such as
223 proper names denote objects of type `e`. Then verb phrases denote
224 functions from individuals to truth values, i.e., functions of type `e
227 The meaning of extraordinary DPs such as *every woman* or *no dog*
228 can't be expressed as a simple individual. As Montague argued, it
229 works much better to view them as predicates on verb phrase meanings,
230 i.e., as having type `(e->t)->t`. Then *no woman left* is true just
231 in case the property of leaving is true of no woman:
233 no woman: \k.not \exists x . (woman x) & kx
235 (no woman) (left) = not \exists x . woman x & left x
237 Montague also proposed that all determiner phrases should have the
238 same type. After all, we can coordinate proper names with
239 quantificational DPs, as in *John and no dog left*. Then generalized
240 quantifier corresponding to the proper name *John* is the quantifier
243 ## How continuations can simulate other monads
245 Because the continuation monad allows the result type ρ to be any
246 type, we can choose ρ in clever ways that allow us to simulate other
250 State: ρ = s -> (α, s)
251 Maybe: ρ = Just α | Nothing
253 You see how this is going to go. Let's see an example by adding an
254 abort operator to our task language, which represents what
255 we want to have happen if we divide by zero, where what we want to do
260 map2 f u v k = u(\u' -> v (\v' -> k(f u' v')))
261 t13 = map2 (++) (mid "a")
266 t13 (\k->Just k) == Just "abcd"
268 t14 = map2 (++) (mid "a")
274 t14 (\k->Just k) == Nothing
278 ## Continuation Passing Style Transforms
280 Gaining control over order of evaluation
281 ----------------------------------------
283 We know that evaluation order matters. We're beginning to learn how
284 to gain some control over order of evaluation (think of Jim's abort handler).
285 We continue to reason about order of evaluation.
287 A lucid discussion of evaluation order in the
288 context of the lambda calculus can be found here:
289 [Sestoft: Demonstrating Lambda Calculus Reduction](http://www.itu.dk/~sestoft/papers/mfps2001-sestoft.pdf).
290 Sestoft also provides a lovely on-line lambda evaluator:
291 [Sestoft: Lambda calculus reduction workbench](http://www.itu.dk/~sestoft/lamreduce/index.html),
292 which allows you to select multiple evaluation strategies,
293 and to see reductions happen step by step.
295 Evaluation order matters
296 ------------------------
298 We've seen this many times. For instance, consider the following
299 reductions. It will be convenient to use the abbreviation `w =
301 indicate which lambda is about to be reduced with a * underneath:
309 Done! We have a normal form. But if we reduce using a different
310 strategy, things go wrong:
324 As a second reminder of when evaluation order matters, consider using
325 `Y = \f.(\h.f(hh))(\h.f(hh))` as a fixed point combinator to define a recursive function:
329 (\f.(\h.f(hh))(\h.f(hh))) (\f n. blah)
331 (\f.f((\h.f(hh))(\h.f(hh)))) (\f n. blah)
333 (\f.f(f((\h.f(hh))(\h.f(hh))))) (\f n. blah)
335 (\f.f(f(f((\h.f(hh))(\h.f(hh)))))) (\f n. blah)
338 And we never get the recursion off the ground.
341 Using a Continuation Passing Style transform to control order of evaluation
342 ---------------------------------------------------------------------------
344 We'll present a technique for controlling evaluation order by transforming a lambda term
345 using a Continuation Passing Style transform (CPS), then we'll explore
346 what the CPS is doing, and how.
348 In order for the CPS to work, we have to adopt a new restriction on
349 beta reduction: beta reduction does not occur underneath a lambda.
350 That is, `(\x.y)z` reduces to `z`, but `\u.(\x.y)z` does not reduce to
351 `\u.z`, because the `\u` protects the redex in the body from
352 reduction. (In this context, a "redex" is a part of a term that matches
353 the pattern `...((\xM)N)...`, i.e., something that can potentially be
354 the target of beta reduction.)
356 Start with a simple form that has two different reduction paths:
358 reducing the leftmost lambda first: `(\x.y)((\x.z)u) ~~> y`
360 reducing the rightmost lambda first: `(\x.y)((\x.z)u) ~~> (\x.y)z ~~> y`
362 After using the following call-by-name CPS transform---and assuming
363 that we never evaluate redexes protected by a lambda---only the first
364 reduction path will be available: we will have gained control over the
365 order in which beta reductions are allowed to be performed.
367 Here's the CPS transform defined:
371 [MN] = \k.[M](\m.m[N]k)
373 Here's the result of applying the transform to our simple example:
376 \k.[\x.y](\m.m[(\x.z)u]k) =
377 \k.(\k.k(\x.[y]))(\m.m(\k.[\x.z](\m.m[u]k))k) =
378 \k.(\k.k(\x.y))(\m.m(\k.(\k.k(\x.z))(\m.muk))k)
380 Because the initial `\k` protects (i.e., takes scope over) the entire
381 transformed term, we can't perform any reductions. In order to watch
382 the computation unfold, we have to apply the transformed term to a
383 trivial continuation, usually the identity function `I = \x.x`.
385 [(\x.y)((\x.z)u)] I =
386 (\k.[\x.y](\m.m[(\x.z)u]k)) I
388 [\x.y](\m.m[(\x.z)u] I) =
389 (\k.k(\x.y))(\m.m[(\x.z)u] I)
391 (\x.y)[(\x.z)u] I --A--
395 The application to `I` unlocks the leftmost functor. Because that
396 functor (`\x.y`) throws away its argument (consider the reduction in the
397 line marked (A)), we never need to expand the
398 CPS transform of the argument. This means that we never bother to
399 reduce redexes inside the argument.
401 Compare with a call-by-value xform:
405 {MN} = \k.{M}(\m.{N}(\n.mnk))
407 This time the reduction unfolds in a different manner:
409 {(\x.y)((\x.z)u)} I =
410 (\k.{\x.y}(\m.{(\x.z)u}(\n.mnk))) I
412 {\x.y}(\m.{(\x.z)u}(\n.mnI)) =
413 (\k.k(\x.{y}))(\m.{(\x.z)u}(\n.mnI))
415 {(\x.z)u}(\n.(\x.{y})nI) =
416 (\k.{\x.z}(\m.{u}(\n.mnk)))(\n.(\x.{y})nI)
418 {\x.z}(\m.{u}(\n.mn(\n.(\x.{y})nI))) =
419 (\k.k(\x.{z}))(\m.{u}(\n.mn(\n.(\x.{y})nI)))
421 {u}(\n.(\x.{z})n(\n.(\x.{y})nI)) =
422 (\k.ku)(\n.(\x.{z})n(\n.(\x.{y})nI))
424 (\x.{z})u(\n.(\x.{y})nI) --A--
427 (\k.kz)(\n.(\x.{y})nI)
436 In this case, the argument does get evaluated: consider the reduction
437 in the line marked (A).
439 Both xforms make the following guarantee: as long as redexes
440 underneath a lambda are never evaluated, there will be at most one
441 reduction available at any step in the evaluation.
442 That is, all choice is removed from the evaluation process.
444 Now let's verify that the CBN CPS avoids the infinite reduction path
445 discussed above (remember that `w = \x.xx`):
448 (\k.[\x.y](\m.m[ww]k)) I
451 (\k.k(\x.y))(\m.m[ww]I)
458 Questions and exercises:
460 1. Prove that {(\x.y)(ww)} does not terminate.
462 2. Why is the CBN xform for variables `[x] = x` instead of something
463 involving kappas (i.e., `k`'s)?
465 3. Write an Ocaml function that takes a lambda term and returns a
466 CPS-xformed lambda term. You can use the following data declaration:
468 type form = Var of char | Abs of char * form | App of form * form;;
470 4. The discussion above talks about the "leftmost" redex, or the
471 "rightmost". But these words apply accurately only in a special set
472 of terms. Characterize the order of evaluation for CBN (likewise, for
473 CBV) more completely and carefully.
475 5. What happens (in terms of evaluation order) when the application
476 rule for CBV CPS is changed to `{MN} = \k.{N}(\n.{M}(\m.mnk))`?
478 6. A term and its CPS xform are different lambda terms. Yet in some
479 sense they "do" the same thing computationally. Make this sense
483 Thinking through the types
484 --------------------------
486 This discussion is based on [Meyer and Wand 1985](http://citeseer.ist.psu.edu/viewdoc/download?doi=10.1.1.44.7943&rep=rep1&type=pdf).
488 Let's say we're working in the simply-typed lambda calculus.
489 Then if the original term is well-typed, the CPS xform will also be
490 well-typed. But what will the type of the transformed term be?
492 The transformed terms all have the form `\k.blah`. The rule for the
493 CBN xform of a variable appears to be an exception, but instead of
494 writing `[x] = x`, we can write `[x] = \k.xk`, which is
495 eta-equivalent. The `k`'s are continuations: functions from something
496 to a result. Let's use σ as the result type. The each `k` in
497 the transform will be a function of type ρ --> σ for some
500 We'll need an ancilliary function ': for any ground type a, a' = a;
501 for functional types a->b, (a->b)' = ((a' -> σ) -> σ) -> (b' -> σ) -> σ.
503 Call by name transform
507 [x] = \k.xk [a] = (a'->o)->o
508 [\xM] = \k.k(\x[M]) [a->b] = ((a->b)'->o)->o
509 [MN] = \k.[M](\m.m[N]k) [b] = (b'->o)->o
511 Remember that types associate to the right. Let's work through the
512 application xform and make sure the types are consistent. We'll have
520 m:((a'->o)->o)->(b'->o)->o
523 [M]:((a->b)'->o)->o = ((((a'->o)->o)->(b'->o)->o)->o)->o
527 Be aware that even though the transform uses the same symbol for the
528 translation of a variable (i.e., `[x] = x`), in general the variable
529 in the transformed term will have a different type than in the source
532 Excercise: what should the function ' be for the CBV xform? Hint:
533 see the Meyer and Wand abstract linked above for the answer.
539 It is easy to think that CBN and CBV are the only two CPS transforms.
540 (We've already seen a variant on call-by-value one of the excercises above.)
542 In fact, the number of distinct transforms is unbounded. For
543 instance, here is a variant of CBV that uses the same types as CBN:
547 <MN> = \k.<M>(\m.<N>(\n.m(\k.kn)k))
549 Try reducing `<(\x.x) ((\y.y) (\z.z))> I` to convince yourself that
550 this is a version of call-by-value.
552 Once we have two evaluation strategies that rely on the same types, we
559 [MN] = \k.<M>(\m.m<N>k)
560 <MN> = \k.[M](\m.[N](\n.m(\k.kn)k))
562 This xform interleaves call-by-name and call-by-value in layers,
563 according to the depth of embedding.
564 (Cf. page 4 of Reynold's 1974 paper ftp://ftp.cs.cmu.edu/user/jcr/reldircont.pdf (equation (4) and the
565 explanation in the paragraph below.)
570 We've already approached some tasks now by programming in **continuation-passing 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 self-consciously when discussing [aborts](/couroutines_and_aborts),
571 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.
573 In our discussion of aborts, we showed how to rearrange code like this:
577 +---try begin----------------+
578 | (if x = 1 then 10 |
581 +---end----------------------+
584 into a form like this:
587 in let snapshot = fun box ->
589 in (foo_result) + 1000
590 in let continue_normally = fun from_value ->
591 let value = from_value + 100
594 if x = 1 then continue_normally 10
598 # #require "delimcc";;
600 # let reset body = let p = new_prompt () in push_prompt p (body p);;
602 let snapshot = fun box ->
604 in (foo_result) + 1000
605 in let continue_normally = fun from_value ->
606 let value = from_value + 100
608 in if x = 1 then continue_normally 10
612 +===try begin================+
613 | (if x = 1 then 10 |
616 +===end======================+
620 let foo x = reset(fun p () ->
622 if x = 1 then k 10 else 20)
636 How did we figure out how to rearrange that code? There are algorithms that can do this for us mechanically. These algorithms are known as **CPS transforms**, because they transform code that might not yet be in CPS form into that form.
638 We won't attempt to give a full CPS transform for OCaml; instead we'll just focus on the lambda calculus and a few extras, to be introduced as we proceed.
640 In fact there are multiple ways to do a CPS transform. Here is one:
643 [\x. M] --> \k. k (\x. [M])
644 [M N] --> \k. [M] (\m. m [N] k)
649 [\x. M] --> \k. k (\x. [M])
650 [M N] --> \k. [M] (\m. [N] (\n. m n k))
652 These transforms have some interesting properties. One is that---assuming we never reduce inside a lambda term, but only when redexes are present in the outermost level---the 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 *call-by-name* 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 *call-by-value* 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.
654 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 [Call-by-name, call-by-value, and the lambda-calculus](http://homepages.inf.ed.ac.uk/gdp/publications/cbn_cbv_lambda.pdf).)
656 Here's another interesting fact about these transforms. Compare the translations for variables and applications in the call-by-value transform:
659 [M N] --> \k. [M] (\m. [N] (\n. m n k))
661 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:
663 let cont_unit x = fun k -> k x
664 let cont_bind N M = fun k -> N (fun n -> M n k)
666 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.) <!-- FIXME -->
668 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 call-by-value CPS transform set out above:
670 let var = \x (\k. k x) in
671 let lam = \x_body (\k. k (\x. x_body x)) in
672 let app = \m n. (\k. m (\m. n (\n. m n k))) in
675 Then if you want to use [x], you'd write `var x`. If you want to use [\x. body], you'd write `lam (\x. BODY)`, where `BODY` is whatever [body] amounts to. If you want to use [m n], you'd write `app M N`, where M and N are whatever [m] and [n] amount to.
677 To play around with this, you'll also want to help yourself to some primitives already in CPS form. (You won't want to rebuild everything again from scratch.) For a unary function like `succ`, you can take its primitive CPS analogue [succ] to be `\u. u (\a k. k (succ a))` (where `succ` in this expansion is the familiar non-CPS form of `succ`). Then for example:
680 = \k. [succ] (\m. [x] (\n. m n k))
684 Or, using the lambda evaluator, that is:
687 let op1 = \op. \u. u (\a k. k (op a)) in
688 app (op1 succ) (var x)
691 Some other handy tools:
693 let app2 = \a b c. app (app a b) c in
694 let app3 = \a b c d. app (app (app a b) c) d in
695 let op2 = \op. \u. u (\a v. v (\b k. k (op a b))) in
696 let op3 = \op. \u. u (\a v. v (\b w. w (\c k. k (op a b c)))) in
699 Then, for instance, [plus x y] would be rendered in the lambda evaluator as:
701 app2 (op2 plus) (var x) (var y)
704 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 continuation-expecting value.
706 If the program you're working with is already in CPS form, then some elegant and powerful computational patterns become available, as we've been seeing. But it's tedious to convert to and work in fully-explicit CPS form. Usually you'll just want to be using the power of continuations at some few points in your program. It'd be nice if we had some way to make use of those patterns without having to convert our code explicitly into CPS form.
711 Well, we can. Consider the space of lambda formulas. Consider their image under a CPS transform. There will be many well-formed lambda expressions not in that image---that is, expressions that aren't anybody's CPS transform. Some of these will be useful levers in the CPS patterns we want to make use of. We can think of them as being the CPS transforms of some new syntax in the original language. For example, the expression `callcc` is explained as a new bit of syntax having some of that otherwise unclaimed CPS real-estate. The meaning of the new syntax can be understood in terms of how the CPS transform we specify for it behaves, when the whole language is in CPS form.
713 I won't give the CPS transform for `callcc` itself, but instead for the complex form:
715 [callcc (\k. body)] = \outk. (\k. [body] outk) (\v localk. outk v)
717 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 application---what is scheduled to happen to `k v` after it's evaluated and `foo` gets access to it---will 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.
719 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.
721 It's important to understand that `callcc` binds `k` to a pipe into the continuation as still then installed. Not just to a function that performs the same computation as the context E[\_] does---that has the same normal form and extension. But rather, a pipe into E[\_] *in its continuation-playing role*. This is manifested by the fact that when `k v` finishes evaluating, that value is not delivered to `foo` for the computation to proceed. Instead, when `k v` finishes evaluating, the program will then be done. Not because of some "stop here" block attached to `k`, but rather because of what it is that `k` represents. Walking through the explanation above several times may help you understand this better.
723 So too will examples. We'll give some examples, and show you how to try them out in a variety of formats:
725 1. using the lambda evaluator to check how the CPS transforms reduce
727 To do this, you can use the following helper function:
729 let callcc = \k_body. \outk. (\k. (k_body k) outk) (\v localk. outk v) in
732 Used like this: [callcc (\k. body)] = `callcc (\k. BODY)`, where `BODY` is [body].
734 2. using a `callcc` operation on our Continuation monad
736 This is implemented like this:
738 let callcc body = fun outk -> body (fun v localk -> outk v) outk
742 -- cutting for control operators --
744 3. `callcc` was originally introduced in Scheme. There it's written `call/cc` and is an abbreviation of `call-with-current-continuation`. Instead of the somewhat bulky form:
746 (call/cc (lambda (k) ...))
748 I prefer instead to use the lighter, and equivalent, shorthand:
753 Callcc/letcc examples
754 ---------------------
756 First, here are two examples in Scheme:
758 (+ 100 (let/cc k (+ 10 1)))
761 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`.
763 What if we do instead:
765 (+ 100 (let/cc k (+ 10 (k 1))))
766 |---------------------|
768 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`.
770 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:
772 (let ([p (let/cc k (cons 1 k))])
773 (cons (car p) ((cdr p) (cons 2 (lambda (x) x)))))
774 ; evaluates to '(2 2 . #<procedure>)
776 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 . #<identity procedure>)`. 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 . #<identity procedure>)`, 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 . #<identity procedure>)`. Racket displays this nested pair like this:
778 '(2 2 . #<procedure>)
782 Ok, so now let's see how to perform these same computations via CPS.
784 In the lambda evaluator:
786 let var = \x (\k. k x) in
787 let lam = \x_body (\k. k (\x. x_body x)) in
788 let app = \m n. (\k. m (\m. n (\n. m n k))) in
789 let app2 = \a b c. app (app a b) c in
790 let app3 = \a b c d. app (app (app a b) c) d in
791 let op1 = \op. \u. u (\a k. k (op a)) in
792 let op2 = \op. \u. u (\a v. v (\b k. k (op a b))) in
793 let op3 = \op. \u. u (\a v. v (\b w. w (\c k. k (op a b c)))) in
794 let callcc = \k_body. \outk. (\k. (k_body k) outk) (\v localk. outk v) in
796 ; (+ 100 (let/cc k (+ 10 1))) ~~> 111
797 app2 (op2 plus) (var hundred) (callcc (\k. app2 (op2 plus) (var ten) (var one)))
798 ; evaluates to \k. k (plus hundred (plus ten one))
802 ; (+ 100 (let/cc k (+ 10 (k 1)))) ~~> 101
803 app2 (op2 plus) (var hundred) (callcc (\k. app2 (op2 plus) (var ten) (app (var k) (var one))))
804 ; evaluates to \k. k (plus hundred one)
806 We won't try to do the third example in this framework.
808 Finally, using the Continuation monad from our OCaml monad library. We begin:
810 # #use "path/to/monads.ml"
811 # module C = Continuation_monad;;
813 Now what we want to do is something like this:
815 # C.(run0 (100 + callcc (fun k -> 10 + 1)));;
817 `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 type-check, 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:
819 # C.(run0 (callcc (fun k -> 10 + 1) >>= fun i -> 100 + i));;
821 Except that's still no good, because `10 + 1` and `100 + i` are of type `int`, but their context demands Continuation monadic values. So we have to throw in some `unit`s:
823 # C.(run0 (callcc (fun k -> unit (10 + 1)) >>= fun i -> unit (100 + i)));;
826 This works and as you can see, delivers the same answer `111` that we got by the other methods.
830 # C.(run0 (callcc (fun k -> unit (10 + (k 1))) >>= fun i -> unit (100 + i)));;
832 That won't work because `k 1` doesn't have type `int`, but we're trying to add it to `10`. So we have to do instead:
834 # C.(run0 (callcc (fun k -> k 1 >>= fun j -> unit (10 + j)) >>= fun i -> unit (100 + i)));;
837 This also works and as you can see, delivers the expected answer `101`.
839 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:
841 # C.(run0 (callcc (fun k -> unit (1,`Box k)) >>= fun (p1,`Box p2) -> p2 (2,`Box unit) >>= fun p2' -> unit (p1,p2')));;
842 - : int * (int * [ `Box of 'b -> ('a, 'b) C.m ] as 'b) as 'a =
847 -- cutting following section for control operators --
849 Some callcc/letcc exercises
850 ---------------------------
852 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.
854 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.
856 Most of the examples assume the following preface:
861 (and (not (pair? x)) (not (null? x))))
863 Now try to figure out what this function does:
867 (let/cc k ; now what will happen when k is called?
868 (letrec ([aux (lambda (l)
871 [(eq? (car l) a) (k (aux (cdr l)))]
872 [else (cons (car l) (aux (cdr l)))]))])
875 Here is [the answer](/hints/cps_hint_1), but try to figure it out for yourself.
877 Next, try to figure out what this function does:
881 (let/cc k ; now what will happen when k is called?
882 (letrec ([aux (lambda (l)
885 [(atom? (car l)) (k (car l))]
887 ; what will the value of the next line be? why is it ignored?
892 Here is [the answer](/hints/cps_hint_2), but try to figure it out for yourself.
894 Next, try to figure out what this function does:
898 (letrec ([aux (lambda (l k)
900 [(null? l) (k 'notfound)]
901 [(eq? (car l) a) (cdr l)]
902 [(atom? (car l)) (cons (car l) (aux (cdr l) k))]
904 ; what happens when (car l) exists but isn't an atom?
905 (let ([car2 (let/cc k2 ; now what will happen when k2 is called?
908 ; when will the following condition be met? what happens then?
909 [(eq? car2 'notfound) (cons (car l) (aux (cdr l) k))]
910 [else (cons car2 (cdr l))]))]))]
911 [lst2 (let/cc k1 ; now what will happen when k1 is called?
914 ; when will the following condition be met?
915 [(eq? lst2 'notfound) lst]
918 Here is [the answer](/hints/cps_hint_3), but try to figure it out for yourself.
920 Here is the hardest example. Try to figure out what this function does:
923 (letrec ([yield (lambda (x) x)]
924 [resume (lambda (x) x)]
927 ; is this the only case where walk returns a non-atom?
929 [(atom? (car l)) (begin
931 (set! resume k2) ; now what will happen when resume is called?
932 ; when the next line is executed, what will yield be bound to?
934 ; when will the next line be executed?
937 ; what will the value of the next line be? why is it ignored?
940 [next (lambda () ; next is a thunk
942 (set! yield k3) ; now what will happen when yield is called?
943 ; when the next line is executed, what will resume be bound to?
945 [check (lambda (prev)
949 [(atom? n) (check n)]
950 ; when will n fail to be an atom?
953 (let ([fst (let/cc k1 (begin
954 (set! yield k1) ; now what will happen when yield is called?
956 ; when will the next line be executed?
959 [(atom? fst) (check fst)]
960 ; when will fst fail to be an atom?
964 Here is [the answer](/hints/cps_hint_4), but again, first try to figure it out for yourself.
967 Delimited control operators
968 ===========================
970 Here again is the CPS transform for `callcc`:
972 [callcc (\k. body)] = \outk. (\k. [body] outk) (\v localk. outk v)
974 `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.
976 (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.)
978 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.
983 ------- reset -------
985 | +----shift k ---+ |
989 | +---------------+ |
991 +-------------------+
994 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.
996 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).
998 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).
1000 For instance, in Scheme this:
1002 (require racket/control)
1006 (+ 10 (shift k x))))
1008 will return 1. The `(let ([x 1]) ...` part is evaluated, but the `(+ 10 ...` part is not.
1010 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.
1012 This pattern should look somewhat familiar. Recall from our discussion of aborts, and repeated at the top of this page:
1015 +---try begin----------------+
1016 | (if x = 1 then 10 |
1019 +---end----------------------+
1022 The box is working like a reset. The `abort` is implemented with a `shift`. Earlier, we refactored our code into a more CPS form:
1025 in let snapshot = fun box ->
1026 let foo_result = box
1027 in (foo_result) + 1000
1028 in let continue_normally = fun from_value ->
1029 let value = from_value + 100
1032 if x = 1 then continue_normally 10
1035 `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.
1041 Using `shift` and `reset` operators in OCaml, this would look like this:
1043 #require "delimcc";;
1044 let p = Delimcc.new_prompt ();;
1045 let reset = Delimcc.push_prompt p;;
1046 let shift = Delimcc.shift p;;
1047 let abort = Delimcc.abort p;;
1051 shift(fun continue ->
1052 if x = 1 then continue 10
1059 If instead at the end we did `... foo 1 + 1000`, we'd get the result `1110`.
1061 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.
1063 You can make the lambda evaluator perform the required CPS transforms with these helper functions:
1065 let reset = \body. \outk. outk (body (\i i)) in
1066 let shift = \k_body. \midk. (\k. (k_body k) (\i i)) (\a localk. localk (midk a)) in
1067 let abort = \body. \midk. body (\i i) in
1070 You use these like so:
1072 * [reset body] is `reset BODY` where `BODY` is [body]
1073 * [shift k body] is `shift (\k. BODY)` where `BODY` is [body]
1074 * and [abort value] is `abort VALUE` where `VALUE` is [value]
1076 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.
1079 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.
1081 (You can start by reading [the Racket docs](http://docs.racket-lang.org/reference/cont.html?q=shift&q=do#%28part._.Classical_.Control_.Operators%29).)
1084 Ken Shan has done terrific work exploring the relations of `shift` and `control` and other control operators to each other.
1086 In collecting these CPS transforms and implementing the monadic versions, we've been helped by Ken and by Oleg and by these papers:
1088 * Danvy and Filinski, "Representing control: a study of the CPS transformation" (1992)
1089 * Sabry, "Note on axiomatizing the semantics of control operators" (1996)
1092 -- cutting some of the following for control operators --
1094 Examples of shift/reset/abort
1095 -----------------------------
1097 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.)
1102 ; (+ 1000 (+ 100 (abort 11))) ~~> 11
1104 app2 (op2 plus) (var thousand)
1105 (app2 (op2 plus) (var hundred) (abort (var eleven)))
1107 # Continuation_monad.(run0(
1108 abort 11 >>= fun i ->
1109 unit (100+i) >>= fun j ->
1113 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:
1115 # Continuation_monad.(run0(
1117 abort 11 >>= fun i ->
1118 unit (100+i) >>= fun j ->
1125 ; (+ 1000 (reset (+ 100 (abort 11)))) ~~> 1011
1127 app2 (op2 plus) (var thousand)
1128 (reset (app2 (op2 plus) (var hundred) (abort (var eleven))))
1130 # Continuation_monad.(run0(
1132 abort 11 >>= fun i ->
1140 ; (+ 1000 (reset (+ 100 (shift k (+ 10 1))))) ~~> 1011
1142 app2 (op2 plus) (var thousand)
1143 (reset (app2 (op2 plus) (var hundred)
1144 (shift (\k. ((op2 plus) (var ten) (var one))))))
1146 Continuation_monad.(
1148 let u = shift (fun k -> unit (10 + 1))
1149 in u >>= fun x -> unit (100 + x)
1150 ) in let w = v >>= fun x -> unit (1000 + x)
1156 ; (+ 1000 (reset (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111
1158 app2 (op2 plus) (var thousand)
1159 (reset (app2 (op2 plus) (var hundred)
1160 (shift (\k. (app (var k) ((op2 plus) (var ten) (var one)))))))
1162 Continuation_monad.(
1164 let u = shift (fun k -> k (10 :: [1]))
1165 in u >>= fun x -> unit (100 :: x)
1166 ) in let w = v >>= fun x -> unit (1000 :: x)
1168 - : int list = [1000; 100; 10; 1]
1170 To demonstrate the different adding order between Examples 4 and 5, we use `::` in the OCaml version instead of `+`. Here is Example 5:
1172 ; (+ 1000 (reset (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently
1174 app2 (op2 plus) (var thousand)
1175 (reset (app2 (op2 plus) (var hundred)
1176 (shift (\k. ((op2 plus) (var ten) (app (var k) (var one)))))))
1178 Continuation_monad.(let v = reset (
1179 let u = shift (fun k -> k [1] >>= fun x -> unit (10 :: x))
1180 in u >>= fun x -> unit (100 :: x)
1181 ) in let w = v >>= fun x -> unit (1000 :: x)
1183 - : int list = [1000; 10; 100; 1]
1188 ; (+ 100 ((reset (+ 10 (shift k k))) 1)) ~~> 111
1190 app2 (op2 plus) (var hundred)
1191 (app (reset (app2 (op2 plus) (var ten)
1192 (shift (\k. (var k))))) (var one))
1194 (* not sure if this example can be typed as-is in OCaml... this is the best I an do at the moment... *)
1196 # type 'x either = Left of (int -> ('x,'x either) Continuation_monad.m) | Right of int;;
1197 # Continuation_monad.(let v = reset (
1198 shift (fun k -> unit (Left k)) >>= fun i -> unit (Right (10+i))
1199 ) in let w = v >>= fun (Left k) ->
1200 k 1 >>= fun (Right i) ->
1206 # type either = Left of (int -> either) | Right of int;;
1207 # let getleft e = match e with Left lft -> lft | Right _ -> failwith "not a Left";;
1208 # let getright e = match e with Right rt -> rt | Left _ -> failwith "not a Right";;
1209 # 100 + getright (let v = reset (fun p () -> Right (10 + shift p (fun k -> Left k))) in getleft v 1);;
1214 ; (+ 100 (reset (+ 10 (shift k (k (k 1)))))) ~~> 121
1216 app2 (op2 plus) (var hundred)
1217 (reset (app2 (op2 plus) (var ten)
1218 (shift (\k. app (var k) (app (var k) (var one))))))
1220 Continuation_monad.(let v = reset (
1221 let u = shift (fun k -> k 1 >>= fun x -> k x)
1222 in u >>= fun x -> unit (10 + x)
1223 ) in let w = v >>= fun x -> unit (100 + x)
1229 print_endline "=== pa_monad's Continuation Tests ============";;
1231 (1, 5 = C.(run0 (unit 1 >>= fun x -> unit (x+4))) );;
1232 (2, 9 = C.(run0 (reset (unit 5 >>= fun x -> unit (x+4)))) );;
1233 (3, 9 = C.(run0 (reset (abort 5 >>= fun y -> unit (y+6)) >>= fun x -> unit (x+4))) );;
1234 (4, 9 = C.(run0 (reset (reset (abort 5 >>= fun y -> unit (y+6))) >>= fun x -> unit (x+4))) );;
1236 let c = reset(abort 5 >>= fun y -> unit (y+6))
1237 in reset(c >>= fun v1 -> abort 7 >>= fun v2 -> unit (v2+10) ) >>= fun x -> unit (x+20))) );;
1239 (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))) );;
1241 (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))) );;
1243 (12, ["a"] = C.(run0 (reset (shift (fun f -> f [] >>= fun t -> unit ("a"::t) ) >>= fun xv -> shift (fun _ -> unit xv)))) );;
1246 (0, 15 = C.(run0 (let f k = k 10 >>= fun v-> unit (v+100) in reset (callcc f >>= fun v -> unit (v+5)))) );;