Unparsers ignore the terminal

[Jens Axel Søgaard]

Hi All,

The output of the following program is:

    '(foo (prop a) (prop b))

I was hoping for:

    '(foo (id a) (prop b))

The problem is that a syntax object representing a symbol

can be both an identifier and a property. 

Since the nonterminal (foo x p) declares that the first x is an identifier

and p is a property, the unparser could pick the correct unparser.

/Jens Axel

#lang racket

(require nanopass/base)

(define id?              identifier?)

(define (unparse-id id)  (list 'id (syntax-e id)))

(define property?            identifier?)

(define (unparse-property p) (list 'prop (syntax-e p)))

(define-language L

  (entry Foo)

  (terminals ((id       (x)) . => . unparse-id)

             ((property (p)) . => . unparse-property))

  (Foo (f)

    (foo x p)))

(unparse-L

 (with-output-language (L Foo)

   `(foo ,#'a ,#'b)))

[Andy Keep]

Yeah, this is definitely a bug.  The unparsers are written to be very permissive, so you can unparse any nonterminal or terminal, however, it is also lazy in that it just applies the top-level unparser to sub-parts instead of using the context to be smarter about it.

I’ve added an issue to both the racket and scheme versions of the project and assigned them to me, since I think these are probably going to be fairly similar changes (though it has been a while since I looked at the racket code, so it may have changed on me :)

Thanks for reporting this, I’ll try to fix this up soon!

-andy:)

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