Re: Macro for replacing a placeholder in an expression

[Maxime Devos]

These macros all sound more complicated than necessary -- on the first 
one, I've sent you a message with sneek:

;; By: Maxime Devos

;; This does not recurse into #(...).
;; Also, such a construct does not nest well, you can't put a 
replace-result-placeholder inside a replace-result-placeholder meaningfully,
;; so I'm wondering why you're doing this, maybe your goal can be accomplished 
more robustly with a different method.
(eval-when (expand load eval)
  (define (replace-placeholder new code) ; <--- recursively transforms code to 
replace '<?>' by new
    (syntax-case code (<?>)
      (<?> new)
      ((x . y)
       #`(#,(replace-placeholder new #'x) . #,(replace-placeholder new #'y)))
      (rest #'rest))))

(define-syntax replace-result-placeholder
  (lambda (s)
    (syntax-case s (<?>) ; <?>: placeholder
      ((_ new code) (replace-placeholder #'new #'code)))))

(display (replace-result-placeholder
           quote
           (<?> bar))) ; -> bar

(I think thinking in terms of 'operations' and special-casing lambda etc 
would make things harder here)

As a bonus, this supports things like `((x . <?>) (z . w)) which aren't 
supported by the original macro as that macro assumed lists.

Greetings,
Maxime.

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