The following puzzles me: #lang racket (define plus +) (free-identifier=? #'+ #'plus) ; -> #f #lang racket (define-syntax (a stx) (syntax-case stx () ((_) (datum->syntax stx (free-identifier=? #'+ #'plus))))) (define plus +) (a) ; -> #f #lang racket (define plus +) (define-syntax (a stx) (syntax-case stx (+) ((_ +) #'#t) ((_ x) #'#f))) (a plus) ; -> #f I am confused, because I expect #t to be produced in the three above cases. Obviously I don't understand free-identifier=? well. Can you help me with that? As another question: the docs on syntax-case state: "An id that has the same binding as a literal-id matches a syntax object that is an identifier with the same binding in the sense of free-identifier=? <file:///C:/Program%20Files/Racket/doc/reference/stxcmp.html?q=syntax-case#% 28def._%28%28quote._~23~25kernel%29._free-identifier~3d~3f%29%29> . The match does not introduce any <file:///C:/Program%20Files/Racket/doc/reference/stx-patterns.html?q=syntax- case#%28tech._pattern._variable%29> pattern variables." Why isn't or cant't the match introduce a pattern variable? Without binding the literal-id as a pattern variable, location information is lost, for example: #lang racket (error-print-source-location #t) (begin-for-syntax (error-print-source-location #t)) (define-syntax (a stx) (syntax-case stx (+) ((_ +) (raise-syntax-error 'a "msg" stx #'+)))) ; <--- (a +) raises a syntax-error as expected, but it highlights + in the line marked <---, not in the last line. I would prefer the + in the last line to be highlighted. (I am running DrRacket, version 6.1.1 [3m]) Help very much appreciated. Wish all of you a fine easter, Jos Koot
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