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Continuation-passing Style

Given (f (g (h i) j) k) what can be done first? => (h i) Why? It must be evaluated before (g(h i)j).

How about (f (g (h i) (j l)))? => either (h i) or (j l).

How to take control?

Start with the following expression:

(hi
  (lambda(hi)
    ...))
  1. Assume that hi is the result of applying (h i)
  2. drop in everything else to replace the ...

so … (f (g (h i) (j l))) becomes

(hi
  (lamda (hi)
    (f (g hi (j l)))

(lambda (hi) (f (g hi (j l)))) is a continuation because:

  1. hi appears in the body of the continuation once
  2. hi is intented to replace (h i) and only (h i)

Write rember* in CPS

In direct style

(define rember8
  (lambda(ls)
    (cond
      [(empty? ls) '()]
      [(eqv? (car ls) 8) (cdr ls)]
      [else (cons (car ls) (rember8 (cdr ls)))])))
  • Serious:

How to CPS expressions:

To convert to CPS style use the following rules

Rule 1: when ever we see a lambda in the code and we want to use CPS style then A. add an argument B. process the body

so (lambda (x ...) ...) => (lambda (x ... k)...^)

Start by added the argument

(define rember8
  (lambda (ls k)
    (cond
      [(null? ls) '()]
      [(eqv? (car ls) 8) (cdr ls)]
      [else (cons (car ls) (rember8 (cdr ls)))])))

Rule 2: Don’t sweat the small stuff!!!! -small stuff is stuff we know will terminate right away -don’t sweat it if you know it will be evaluated -if it might be evaluated, instead pass it to k

Ex. (empty? ls) -we know it will be evaluated -we know it is small stuff -we don’t worry about it

So what do you do with the '() that is returned as the answer to (empty? ls)? => pass it to k

(define rember8
  (lambda(ls k)
    (cond
      [(empty? ls) (k '()) ] ;;small stuff
      [(eqv? (car ls) 8) (k (cdr ls))] ;;small stuff
      [else (cons (car ls) (rember8 (cdr ls)))] ;; not small stuff
      )))

[else (cons (car ls) (rember8 (cdr ls)))] is not small stuff so need to build a new continuation there is still small stuff in the body so we pass it to k the else line is shown on the following

Tail position

Non-tail position

Caution about tail position

;rember8 that is CPSed
(define rember8
  (lambda (ls k)
    (cond
      [(empty? ls) (k '()) ]
      [(eqv? (car ls) 8) (k (cdr ls))] 
      [else (rember8 (cdr ls)
                       (lambda (x)
                         (k
                          (cons (car ls) x))))]
      )))

how do you invoke it? => we need a k and ls to pass in as follows (rember8 '() k) => '()

k could be the identify function (lambda (x) x)

(rember8 '(1 2 8 3 4 6 7 8 5) (lambda (x) x))

What properties can be observed about this program?

  1. All non-small stuff calls are tail calls EX surround tail calls with * here.
(define rember8
  (lambda (ls *k*)
    (cond
      [(empty? ls) (*k* '()) ]
      [(eqv?(car ls) 8) (*k* (cdr ls))] 
      [else (*rember8* (cdr ls)
                       (lambda (x)
                         (*k*
                          (cons (car ls) x))))]
      )))

Simple vs Serious

Why don’t empty?, eqv?, car, cdr, and cons count?

  1. they are small stuff (if we combine small stuff together in small things the combination remains small)
  2. All arguments = small stuff lambda = small stuff

This is essentially a ‘C’ program. Just convert the continuation to data structures. (We did this with closures)

Simple expressions

Serious expressions

  • We decide

How about we trace (rember8 '(1 2 8 3 4 6 7 8 5) (lambda (x) x))

(ls | k)

'(1 2 8 3 4 6 7 8 5) |  (lambda (x) x) = id
'(2 8 3 4 6 7 8 5)   |  (lambda (x)
                          (id (cons 1 x))) = k2
'(8 3 4 6 7 8 5)     |  (lambda (x)
                          (k2 (cons 2 x))) = k3
...

Once we hit 8 we apply (k (cdr ls))

k = k3
ls = '(8 3 4 6 7 8 5)
'(8 3 4 6 7 8 5)     |  (lambda (x)
                          (k2 (cons 2 x))) = k3
(k3 '(3 4 6 7 8 5)) =>
(k2 (cons 2 '(3 4 6 7 8 5))) =>
(id (cons 1 '(2 3 4 6 7 8 5))) =>
(id '(1 2 3 4 6 7 8 5)) => '(1 2 3 4 6 7 8 5)

Done

Another example

;; remove all 8's
(define multirember8
  (lambda (ls)
    (cond
      [(empty? ls) '()]
      [(eqv? (car ls) 8) (multirember8 (cdr ls))]
      [else (cons (car ls) (multirember8 (cdr ls)))])))

Let’s CPS this thing

(define multirember8
  (lambda (ls k)
    (cond
      [(empty? ls) (k '())]
      [(eqv? (car ls) 8) (multirember8 (cdr ls))] ;; **awe snap 
      [else (multirember8 (cdr ls)
                          (lambda (x)
                            (k (cons (car ls) x))))])))

multirember8 takes two arguments => so you need to make another continuation

(define multirember8
  (lambda (ls k)
    (cond
      [(empty? ls) (k '())]
      [(eqv? (car ls) 8) (multirember8 (cdr ls)
                                    (lambda (x)
                                      (k x)))] 
      [else (multirember8 (cdr ls)
                          (lambda (x)
                            (k (cons (car ls) x))))])))

what does (lambda (x) (k x)) do?

=> it takes the whatever is passed “x” and passes it to k

Sooo …

Eta reduction: (lambda (x) (M x)) should go to M if x is not free in M & M is going to terminate

M is any arbitrary expression that satisfies the above rule (does not have to be a single variable like k)

so when every you see a tail call you dont even need to think about Eta => just pass k to it

(define multirember8
  (lambda (ls k)
    (cond
      [(empty? ls) (k '())]
      [(eqv? (car ls) 8) (multirember8 (cdr ls)
                                    k)]
      [else (multirember8 (cdr ls)
                          (lambda (x)
                            (k (cons (car ls) x))))])))

Special/tricky cases

let

cond

Example

(define !
  (lambda (n)
    (if (zero? n)
    1
    (* n (! (sub1 n))))))

You try

(define extend-env
  (lambda (x a env)
    (lambda (y)
  (if (eqv? x y)
      a
      (apply-env env y)))))

(define apply-env
  (lambda (env y)
    (env y)))

Acks

Much of the preceding was borrowed from notes by Adam Foltzer.

Updated: