Why does syntax-parser work here but not syntax-parse?

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David Storrs

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Sep 8, 2021, 1:13:03 PMSep 8
to Racket Users
This is from the documentation and it obviously works:

(define parser1                
  (syntax-parser
    [((~alt (~once (~seq #:a x) #:name "#:a keyword")
            (~optional (~seq #:b y) #:name "#:b keyword")
            (~seq #:c z)) ...)
     'ok]))
(parser1 #'(#:a 1))

When run it yields 'ok.

If I change it to this, it fails and I don't understand why:

(define (parser2 stx)
  (syntax-parse stx
    [(parser2 ((~alt (~once (~seq #:a x) #:name "#:a keyword")
                     (~optional (~seq #:b y) #:name "#:b keyword")
                     (~seq #:c z)) ...))
     #''ok]))
(parser2 (#:a 1))

This yields:
[...source location...] #%datum: keyword misused as an expression
;   at: #:a
                                                        


I can see that (#:a 1) is not valid under the default parser since #:a is not valid for initial position but shouldn't the entire parenthesized expression be given to the macro processor and then replaced with something valid before being rejected?

Stephen Chang

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Sep 8, 2021, 2:03:03 PMSep 8
to David Storrs, Racket Users
> shouldn't the entire parenthesized expression be given to the macro processor and then replaced with something valid before being rejected?

That would be true if you're defining a macro, i.e. if you use
`define-syntax` instead of `define`.

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David Storrs

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Sep 8, 2021, 3:02:14 PMSep 8
to Stephen Chang, Racket Users
*headdesk headdesk headdesk headdesk headdesk*

Thank you.
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