Re: Macro for replacing a placeholder in an expression

[Zelphir Kaltstahl <zelphirkaltstahl@posteo.de>]

Hello Maxime!

On 7/28/22 02:55, Maxime Devos wrote:
>
> 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.
>
I'll need to look at this and learn about eval-when. I also did not think about 
vectors yet. Thank you!

Best regards,
Zelphir

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