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From: Maxime Devos <maximedevos@telenet.be>
To: Zelphir Kaltstahl <zelphirkaltstahl@posteo.de>
Cc: Guile User <guile-user@gnu.org>
Subject: Re: Macro for replacing a placeholder in an expression
Date: Sat, 30 Jul 2022 22:44:43 +0200	[thread overview]
Message-ID: <64ef2c56-2999-9514-c98b-7a14c9683e1a@telenet.be> (raw)
In-Reply-To: <5f0efafc-3045-e710-6066-9519692ebb44@posteo.de>


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On 30-07-2022 17:42, Zelphir Kaltstahl wrote:
>
> [...]
>
> But now comes the problem:
>
> Since I want to replace all occurrences of for example <?> and <?> 
> does not need to be defined, I think I must use define-syntax, to 
> avoid Guile trying to evaluate the arguments to a function call. OK, 
> so a macro I write:
>
> ~~~~
> (define-syntax replace-placeholder
>    (λ (stx)
>      (syntax-case stx (<?>)
>        [(_ replacement <?>)
>         (syntax replacement)]
>        [(_ replacement (car-elem . cdr-elem))
>         (quasisyntax
>          ((unsyntax (replace-placeholder #'replacement #'car-elem)) .
>           (unsyntax (replace-placeholder #'replacement #'cdr-elem))))]
>        [(_ replacement other)
>         (syntax other)])))
> ~~~~
>
> [...]
>
> When I use this on a trivial expression, it works:
>
> ~~~~
> (replace-placeholder 3 <?>)
> => 3
> ~~~~
>
> When I try to use this for a pair as follows:
>
> ~~~~
> (replace-placeholder 3 (+ 1 <?>))
> => While compiling expression:
> Wrong type to apply: #<syntax-transformer replace-placeholder>
> ~~~~
>
> It does not work. What happens here, I guess, is, that the macro gets 
> expanded, then the syntax-transformer ends up in a place like 
> (replace-placeholder …) and since it is not a function, it cannot be 
> applied.
>
I think so to -- syntax isn't procedure.
>
> But this is exactly what I want! I want Guile to do another macro call 
> right there and replace in the sub-expression. How can I tell Guile to 
> do that?
>
To use replace-placeholder as a procedure, you can simply turn it into a 
procedure, by replacing define-syntax with define. Now, because in the 
end you want syntax and not just a procedure, you also define a small 
wrapper using define-syntax.  I expect you will end up with something 
similar to the 'replace-placeholder + replace-result-placeholder' 
example I sent previously.  If you really want to, there is is also the 
'macro-transformer' procedure. If you don't like a separate helper 
procedure (maybe in an eval-when) defined outside the define-syntax, 
there are some tricks to avoid that if you are interested?
>
> I think that only now I am understanding properly what you wrote: 
> "Also, such a construct does not nest well, you can't put a 
> replace-result-placeholder inside a replace-result-placeholder 
> meaningfully, […]". Does this mean, that recursive application of a 
> macro inside a macro is impossible? To expand to subforms being the 
> same macro again and this way transform a whole tree of s-expressions?
>
No, this is not what I meant. What I meant is that things like the 
following won't work well:

(define (plus-one x)
   (replace-result-placeholder x
     (+ <?> (replace-result-placeholder 1 <?>))))

-- if I read this, I would expect it to be equivalent to (lambda (x) (+ 
x 1)), but IIUC, both the innermost and outermost <?> will be replaced 
by x so you end up with (lambda (x) (+ x x)) instead (unverified).


> "All I want to do" is to replace some placeholder (in this case <?>) 
> in an arbitrary form. No matter how that form looks or how deeply it 
> is nested, if there are <?> inside of it, I want to replace them. Is 
> this impossible?
>
Yes, see e.g. the replace-placeholder+replace-result-placeholder I sent, 
subject to the limitations of messy nesting semantics. However ...
>
> Ultimately this is a sub-problem of a bigger thing I want to do. Part 
> of the contracts thingy. I want to make it so, that the following is 
> valid and works:
>
> ~~~~
> (define-with-contract account-withdraw
>    (require (<= amount account-balance)
>             (>= amount 0))
>    (ensure (>= <?> 0)
>            arbitrarily-complex-expression-here-where-placeholder-will-be-replaced-with-function-result-identifier)
>    (λ (amount account-balance)
>      (- account-balance amount)))
> ~~~~
>
> In SRFI 197 someone seems to have done that: 
> https://srfi.schemers.org/srfi-197/srfi-197.html The underscore _ can 
> be anywhere and the result of previous chain steps will be put there.
>
> Perhaps I have to check how that is implemented
>
..., while I'm not familiar with SRFI 197, I would doubt that that SRFI 
does this in __all__ contexts -- I would expect it to keep (quote _) 
intact (unverified, maybe SRFI actually _does_ change that?).

If you want to _not_ change (quote _), try defining <?> as a syntax 
parameter (see (guile)Syntax Parameters) and using syntax-parameterize 
-- if so, you can implement your thing with only syntax-rules and not 
syntax-case (maybe the nesting limitation would be solved too, but I 
don't actually know that for a fact).

Greetings,
Maxime.

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  reply	other threads:[~2022-07-30 20:44 UTC|newest]

Thread overview: 14+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2022-07-27 23:57 Macro for replacing a placeholder in an expression Zelphir Kaltstahl
2022-07-28  0:55 ` Maxime Devos
2022-07-28 10:23   ` Zelphir Kaltstahl
2022-07-28 10:28     ` Maxime Devos
2022-07-30 15:42     ` Zelphir Kaltstahl
2022-07-30 20:44       ` Maxime Devos [this message]
2022-07-30 21:10         ` Maxime Devos
2022-07-30 21:13       ` Maxime Devos
2022-08-05  9:42         ` Linus Björnstam
2022-08-06 14:28           ` Zelphir Kaltstahl
2022-07-28  1:04 ` Maxime Devos
2022-07-28  8:39   ` Zelphir Kaltstahl
2022-07-28  9:48     ` Maxime Devos
2022-07-28 11:26       ` Zelphir Kaltstahl

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