* [Example of an not-fully-immersive game](http://www.i-mockery.com/minimocks/50arcadecabinets/star-wars-arcade.gif)
* [A more immersive game](http://upload.wikimedia.org/wikipedia/commons/7/78/AC89-0437-20_a.jpeg)
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:
(call/cc (lambda (k) ...))
I prefer instead to use the lighter, and equivalent, shorthand:
(let/cc k ...)
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`, then computes `(+ 10 1)` and delivers that to `outk` in the normal way (not through `k`). No unusual behavior. It evaluates to `111`.
What if we do instead:
(+ 100 (let/cc k (+ 10 (k 1))))
|---------------------|
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:
(let ([p (let/cc k (cons 1 k))])
(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 nested pair, whose first element is `2` and whose second element is the pair `'(2 . #)`. Racket displays this nested pair like this:
'(2 2 . #)
---
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.
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.
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 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
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