Because both syntax objects encapsulating a list and a list of syntax objects
are valid syntax objects.
If you match a list using just a pattern var it will be a syntax objects. I'd
you match it using (pattern ...) you will do #'(pattern ...) and get a list.
You can easily implement syntax->list that handles both cases and always
returns a list.
--
Linus Björnstam
On Tue, 5 Apr 2022, at 01:00, Jean Abou Samra wrote:
> Hi,
>
> I am lost as to when a syntax object whose syntax->datum is a pair
> can be manipulated as a plain Scheme pair. For example:
>
>
> (define synt1 #'(a b))
> (pk 'syntax synt1 'pair? (pair? synt1))
> (pk 'car (car synt1))
>
> (define-syntax mysyntax
> (lambda (stax)
> (syntax-case stax ()
> ((_ thing)
> (begin
> (pk 'syntax #'thing 'pair? (pair? #'thing))
> (pk 'car (car #'thing)))))))
>
> (mysyntax (c d))
>
>
> results in
>
>
> ;;; (syntax (#<syntax a> #<syntax b>) pair? #t)
>
> ;;; (car #<syntax a>)
>
> ;;; (syntax #<syntax:test.scm:13:10 (c d)> pair? #f)
> Backtrace:
> 7 (primitive-load "/home/jean/repos/lilypond/build/test.s…")
> In ice-9/eval.scm:
> 721:20 6 (primitive-eval (mysyntax (c d)))
> In ice-9/psyntax.scm:
> 1229:36 5 (expand-top-sequence (#<syntax:test.scm:13:0 (mysynta…>) …)
> 1121:20 4 (parse _ (("placeholder" placeholder)) ((top) #(# # …)) …)
> 1342:32 3 (syntax-type (mysyntax #<syntax:test.scm:13:10 (c d)>) # …)
> 1562:32 2 (expand-macro #<procedure 7fa6fa335320 at ice-9/eval.s…> …)
> In ice-9/eval.scm:
> 159:9 1 (_ #(#(#<directory (guile-user) 7fa6fa3dfc80>) #<synt…>))
> 155:9 0 (_ _)
>
> ice-9/eval.scm:155:9: In procedure car: Wrong type argument in position
> 1 (expecting pair): #<syntax:test.scm:13:10 (c d)>
>
>
> Why does one syntax object look like a pair and not the other?
> What is the difference between the two cases?
>
> Thanks,
> Jean
