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
