string-unfold demo

string-unfold demo

[Matt Wette]

Just for kicks, to learn string-unfold, I made an ugly version of string-append:

(define (ugly-string-append . str-l)

  ;; p: seed |-> #t|#f   predicate to indicate stop
  (define (p seed) (null? seed))

  ;; f: seed |-> char    output function
  (define (f seed) (string-ref (cddar seed) (caar seed)))

  ;; g: seed |-> seed    transition function
  (define (g seed) (let* ((head (car seed)) (tail (cdr seed))
			  (ix (car head)) (ln (cadr head)) (st (cddr head)))
		     (if (eq? (1+ ix) ln) tail
			 (cons (cons* (1+ ix) ln st) tail))))

  ;; s: seed = ((ix1 ln1 . st1) (ix2 ln2 . st2) ...)
  ;;           where ix is curr index, ln is string-length, and st is string
  (define s (map (lambda (s) (cons* 0 (string-length s) s)) str-l))

  (string-unfold p f g s))

(ugly-string-append "abc" "def") => "abcdef"

Re: string-unfold demo

[Alex Vong]

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Hello Guilers,

Since I find this exercise interesting, I come up with another demo. The
trick is to think of string as a list of characters (like in Haskell)
and to use the fact that append can be written as an unfold.

Let's begin:

1. Use SRFI-1 and SRFI-26

  (use-modules (srfi srfi-1)
               (srfi srfi-26))

2. Implement append using unfold

  (define (append a b)
    (unfold null?
            car
            cdr
            a
            (const b)))

Test it:

  (append '(1 2 3) '(4 5 6))
  => (1 2 3 4 5 6)

3. Extend append to allow any number of arguments

  (define append
    (lambda args
      (define (%append a b)
        (unfold null?
                car
                cdr
                a
                (const b)))
      (fold-right %append '() args)))

Test it:

  (append '(1 2 3) '(4 5 6) '(7 8 9))
  => (1 2 3 4 5 6 7 8 9)

4. "Port" the procedure to "string"-land

First, notice 'unfold' takes a TAIL-GEN procedure to generate the tail,
while string-unfold takes a BASE string, not a procedure. So, let's
replace '(const b)' with 'b'. Next, replace 'null?' with 'string-null?',
'car' with '(cut string-ref <> 0)' and 'cdr' with
'(cut string-drop <> 1)'

  (define (string-append a b)
    (string-unfold string-null?
                   (cut string-ref <> 0)
                   (cut string-drop <> 1)
                   a
                   b))

Test it:

  (string-append "123" "456")
  => "456123"

Oops, we got the a and b reversed.

5. Reverse a and b

  (define (string-append a b)
    (string-unfold string-null?
                   (cut string-ref <> 0)
                   (cut string-drop <> 1)
                   b
                   a))

Test it:

  (string-append "123" "456")
  => "123456"

Now it works.

6. Extend again to allow any number of arguments

  (define string-append
    (lambda args
      (define (%string-append a b)
        (string-unfold string-null?
                       (cut string-ref <> 0)
                       (cut string-drop <> 1)
                       b
                       a))
      (fold-right %string-append "" args)))

Test it:

  (string-append "123" "456" "789")
  => "123456789"

DONE!

Cheers,
Alex

Matt Wette <matt.wette@gmail.com> writes:

> Just for kicks, to learn string-unfold, I made an ugly version of string-append:
>
> (define (ugly-string-append . str-l)
>
>   ;; p: seed |-> #t|#f   predicate to indicate stop
>   (define (p seed) (null? seed))
>
>   ;; f: seed |-> char    output function
>   (define (f seed) (string-ref (cddar seed) (caar seed)))
>
>   ;; g: seed |-> seed    transition function
>   (define (g seed) (let* ((head (car seed)) (tail (cdr seed))
> 			  (ix (car head)) (ln (cadr head)) (st (cddr head)))
> 		     (if (eq? (1+ ix) ln) tail
> 			 (cons (cons* (1+ ix) ln st) tail))))
>
>   ;; s: seed = ((ix1 ln1 . st1) (ix2 ln2 . st2) ...)
>   ;;           where ix is curr index, ln is string-length, and st is string
>   (define s (map (lambda (s) (cons* 0 (string-length s) s)) str-l))
>
>   (string-unfold p f g s))
>
> (ugly-string-append "abc" "def") => "abcdef"

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