Re: Understanding `symbol??` macro from okmij.org

[Zelphir Kaltstahl]

On 3/14/24 02:01, Jean Abou Samra wrote:
> Hi!
>
> The explanations are on
>
> https://okmij.org/ftp/Scheme/macros.html#macro-symbol-p
>
> Maybe this version will be easier to understand (we don't really
> need continuation-passing style here):
>
> (define-syntax symbol??
>    (syntax-rules ()
>      ((symbol?? maybe-symbol)
>       (let-syntax
>           ((test
>             (syntax-rules ()
>               ((test maybe-symbol) #t)
>               ((test _) #f))))
>         (test abracadabra)))))
>
> (symbol?? foo) ⇒ #t
> (symbol?? (a . b)) ⇒ #f
> (symbol?? 5) ⇒ #f
> (symbol?? "a") ⇒ #f
> (symbol?? #(1 a)) ⇒ #f
>
>
> Basically: the macro call
>
> (symbol?? <foobar>)
>
> expands to a macro definition of test as
>
> (syntax-rules ()
>    ((test <foobar>) #t)
>    ((test _) #f))
>
> and a call (test abracadabra). Now, observe that if <foobar>
> is a symbol, then it's a catch-all pattern when inserted
> in the syntax-rules definition of `test`, so it will match
> abracadabra (because it matches anything). On the other hand,
> if it's not a symbol, then it won't match abracadabra, by case
> analysis: if it's a number it will only match that number; booleans,
> strings and characters likewise; if it's a pair it can only match
> pairs; if it's a vector it can only match vectors; etc.
>
> I'm not exactly sure why Oleg Kiselyov included special cases for a pair
> and a vector, but my guess is that not all Scheme implementations
> support vectors in syntax-rules patterns (and the pair check is
> necessary because the car or cdr could contain a vector). The Scheme
> standards certainly have their opinion on this, and I knew that
> stuff by heart at some point (when I implemented a syntax-rules/syntax-case
> expander for a university project), but I don't remember, and it's
> too late for scouring the standards...
>
> Best,
> Jean


On 3/14/24 02:03, Maxime Devos wrote:
> (define-syntax symbol??
>
>    (syntax-rules ()
>
>      ;; The check is done by first pattern matching against some other
>
>      ;; forms, that are not a symbol.
>
>      ((symbol?? (x . y) kt kf) kf) ; It's a pair, not a symbol
>
>      ((symbol?? #(x ...) kt kf) kf) ; It's a vector, not a symbol
>
> The first two cases don’t need any explanation I think.
>
>      ;; After those things are excluded, the thing might be a symbol.
>
>      ((symbol?? maybe-symbol kt kf)
>
>       (let-syntax ((test
>
>                     (syntax-rules ()
>
>                       ((test maybe-symbol t f) t)
>
>                       ((test x t f) f))))
>
>         (test abracadabra kt kf)))))
>
> If maybe-symbol is a symbol, then (test abracadabra kt kf) matches (test 
> maybe-symbol t f) (let t=kt, f=kf, maybe-symbol=abracadabra).
>
> If maybe-symbol is not a symbol, for example it is a string “hello”  (please 
> ignore wrong quoting), then
>
> (test maybe-symbol t f) becomes (test “hello” t f).  The string “hello” cannot 
> act as an identifier (because it is a string, not a symbol), so (test 
> abracadabra kt kf) does not match the first case (test maybe-symbol t f). Only 
> the second case (test x t f) remains, and (test abracadabra kt kf) matches 
> this (set x to abracadabra, t to kt, f to kf).
>
> I think the first two cases are superfluous, but perhaps there is a 
> performance advantage.
>
> (I’m wondering if this still works in the case (symbol? ...), because ... is 
> special in syntax-rules)
>
> Best regards,
>
> Maxime Devos.
Thank you both!

Mind-blowing stuff. But I think I understand it now. How clever.

Best regards,
Zelphir

--
repositories:https://notabug.org/ZelphirKaltstahl