Re: Syntax-Case macro that selects the N-th element from a list

[Tim Van den Langenbergh]

On Monday, 5 April 2021 13:30:21 CEST you wrote:
> Hi,
> 
> In dryads-wake I need selection of the element in a list in a macro from
> user-input. Currently I have multiple macros, and the correct one (which
> strips the non-selected choices) is selected in a simple cond:
> 
> (define-syntax-rule (Choose resp . choices)
>    "Ask questions, apply consequences"
>    (cond
>     ((equal? resp 1) ;; resp is user-input. It is a natural number.
>      (Respond1 choices))
>     ((equal? resp 2)
>      (Respond2 choices))
>     ((equal? resp 3)
>      (Respond3 choices))
>     (else
>      #f)))
> 
> For this however I have three syntax-case macros:
> 
> (define-syntax Respond1
>   (lambda (x)
>     (syntax-case x ()
>       ((_ ((question consequences ...) choices ...))
>         #`(begin
>            (respond consequences ...)))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> (define-syntax Respond2
>   (lambda (x)
>     (syntax-case x ()
>       ((_ (choice choices ...))
>         #`(begin
>            (Respond1 (choices ...))))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> (define-syntax Respond3
>   (lambda (x)
>     (syntax-case x ()
>       ((_ (a b choices ...))
>         #`(Respond1 (choices ...)))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> 
> I would like to get rid of those three definitions and replace them by
> at most two (one that strips N initial list entries, and Respond1).
> 
> I cannot move to procedures, because I have code that must be executed
> only during final processing, and when I evaluate any of the
> consequences (as it happens with procedure-arguments), then the timing
> of the code execution does not match anymore. So I must absolutely do
> this in macros.
> 
> 
> I’ve tried to get that working, but all my tries failed. Is there a way
> and can you show it to me?
> 
> This is a minimal working example. The output should stay the same,
> except for part 4, which needs this change to work (see at the bottom),
> but I would like to:
> 
> - replace Respond2 and Respond3 by something recursive, so resp can have
>   arbitrary high values (not infinite: max the length of the options) and
> - replace the cond-clause by a call to the recursive macro.
> 
> (define-syntax-rule (respond consequence consequence2 ...)
>   (begin
>     (write consequence)
>     (when (not (null? '(consequence2 ...)))
>       (write (car (cdr (car `(consequence2 ...))))))))
> 
> (define-syntax Respond1
>   (lambda (x)
>     (syntax-case x ()
>       ((_ ((question consequences ...) choices ...))
>         #`(begin
>            (respond consequences ...)))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> (define-syntax Respond2
>   (lambda (x)
>     (syntax-case x ()
>       ((_ (choice choices ...))
>         #`(begin
>            (Respond1 (choices ...))))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> (define-syntax Respond3
>   (lambda (x)
>     (syntax-case x ()
>       ((_ (a b choices ...))
>         #`(Respond1 (choices ...)))
>       ((_ (choices ...))
>         #`(begin #f)))))
> 
> 
> (define-syntax-rule (Choose resp . choices)
>    "Ask questions, apply consequences"
>    (cond
>     ((equal? resp 1)
>      (Respond1 choices))
>     ((equal? resp 2)
>      (Respond2 choices))
>     ((equal? resp 3)
>      (Respond3 choices))
>     (else
>      #f)))
> 
> 
> (display "Choose 1: should be bar:")
> (Choose 1 (foo 'bar) (foo 'war 'har) (foo 'mar) (foo 'tar))
> (newline)
> (display "Choose 2: should be warhar:")
> (Choose 2 (foo 'bar) (foo 'war 'har) (foo 'mar) (foo 'tar))
> (newline)
> (display "Choose 3: should be mar:")
> (Choose 3 (foo 'bar) (foo 'war 'har) (foo 'mar) (foo 'tar))
> (newline)
> (display "Choose 4: should be tar:")
> (Choose 4 (foo 'bar) (foo 'war 'har) (foo 'mar) (foo 'tar))
> (newline)
> (display "Choose 5: should be #f:")
> (Choose 5 (foo 'bar) (foo 'war 'har) (foo 'mar) (foo 'tar))
> (newline)
> 
> 
> Best wishes,
> Arne

Hello, Dr. Arne,

would simply transforming the question . response pairs to a list of lists of 
responses work?

E.G.

#+begin_src scheme
(define-syntax questions->responses
  (syntax-rules ()
        ((_ (question response ...) choices ...)
         (cons (list response ...)
                   (questions->responses choices ...)))
        ((_ choices ...)
         '())))
#+end_src

Vale,

-Tim