[plt-scheme] sicp exercise 2.23

I have read up to chapter 4 of htdp , so this is how i will do it at first try

;;for-each1

;;for-each1 : (X -> Y) (listof X) -> true
;;to apply a (f X) on items of a-list , on success returns true
(define (for-each1 f a-lst)
  (cond
    [(empty? a-lst) true]
    [else (and (always-true (f (first a-lst))) (for-each1 f (rest a-lst)))]))

;;always-true : X -> true
;;to produce to true for any item
(define (always-true item)
  true)

;;print-item : X -> X
;;to print a-item
(define (print-item a-item)
  a-item)

(for-each1 print-item (list 1 2 3))


On Tue, Apr 29, 2008 at 11:34 AM, Jason Wang <randomtalk at gmail.com> wrote:
> Hey guys, i was just working through SICP book after having (pretty
>  much) completed HtDP in school. I have  a question about 2.23 exercise
>  reproduced here:
>
>  Exercise 2.23.  The procedure for-each is similar to map. It takes as
>  arguments a procedure and a list of elements. However, rather than
>  forming a list of the results, for-each just applies the procedure to
>  each of the elements in turn, from left to right. The values returned
>  by applying the procedure to the elements are not used at all --
>  for-each is used with procedures that perform an action, such as
>  printing. For example,
>
>  (for-each (lambda (x) (newline) (display x))
>           (list 57 321 88))
>  57
>  321
>  88
>
>  The value returned by the call to for-each (not illustrated above) can
>  be something arbitrary, such as true. Give an implementation of
>  for-each.
>  -----------------
>  Using begin, this problem is quite simple, as below:
>  (define (for-each fun items)
>   (cond [(not (null? items))
>          (begin
>            (fun (car items))
>            (for-each fun (cdr items)))]))
>  (for-each (lambda (x) (newline) (display x))
>           (list 57 321 88))
>
>  However, my problem is that SICP hasn't introduced begin yet, so i
>  looked around online and saw someone else's solution as follows:
>  (define (for-each2 fun items)
>   (cond [(not (null? items))
>          (fun (car items))
>          (for-each2 fun (cdr items))]))
>  (for-each2 (lambda (x) (newline) (display x))
>           (list 57 321 88))
>
>  It works as well, but can someone tell me why it works and if i did
>  the following (which i thought is equivalent to the second one) and it
>  wouldn't work:
>  (define (for-each3 fun items)
>   (if (not (null? items))
>       (fun (car items))
>       (for-each3 fun (cdr items))))
>  (for-each3 (lambda (x) (newline) (display x))
>           (list 57 321 88))
>
>  Also why do we learn begin if one of the last 2 works.
>
>  On a side note, the test code is provided in the book, and in the
>  lambda expression, the author didn't use begin either....
>
>  Thanks a lot for any clarifications provided :D
>
>  Jason
>
>  --
>  "It's a wonderful world hobbes ol' buddy, let's go explorin'!" -
>  Calvin in "Calvin and Hobbes"
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