Re: define-module, #:export and export

[Jean Abou Samra]

Le 04/01/2023 à 18:28, Maxime Devos a écrit :
> On 04-01-2023 16:11, yarl baudig wrote:
> > Hello guile.
> >
> > I don't know if that's a bug. Anyway, I am confused about this so I 
> > ask. I came across this problem playing with guix source code. I will 
> > share different "tests" each test is a directory with nothing but the 
> > files I share.
> > each time the command to try the test (inside it's directory) is 
> > `guile --no-auto-compile -L . main.scm`
> > [...]
> My (untested) hypothesis (*: things I'm unsure about)
>
> If there is no #:export (...), then when
> (define-operation (valid-path?)) is expanded, the identifier 
> 'valid-path?' is free (*).
>
> Hence, the following ...
>
> >         (syntax-rules (name ...)
> >           ((_ name) id) ...)))))
>
> refers to the free identifier 'valid-path?'.
>
> Conversely, if there is #:export, then the (syntax-rules (name ...) 
> ...) looks for the bound identifier 'valid-path?' (*).
>
> Important: bound identifiers != free-identifier, even if they have the 
> same symbol!  For (simple-format #t "~S\n" (operation-id valid-path?)) 
> to work, the 'valid-path?' of that expression must be bound if the 
> syntax-rules looks for a bound identifier, and free if it looks for a 
> free identifier.


That is also my understanding, confirmed by

$ cat lib.scm
(define-module (lib)
  #:export (test))

(define-syntax test
  (lambda (sintax)
    (syntax-case sintax ()
      ((test id)
       (datum->syntax sintax (free-identifier=? #'id #'thing))))))

$ cat test.scm
(define-module (main)
  #:use-module (lib)
  #:export (thing)
  )

(display (test thing))
(newline)

(define thing 5)

$ guile3.0 -L . test.scm
#f


If you comment out #:export (thing), the result changes to #t.

To put it perhaps more simply, the use of #:export causes Guile to 
understand early that there will be a variable 'thing' in this module, 
and makes the identifier 'thing' refer to this variable that is not yet 
defined. However, hygiene implies that you want to be able to use 
keywords as if they were not keywords if they are rebound, e.g. the 
'else' here doesn't cause the cond clause to be taken:

(let ((else #f)) (cond (#f 'bla) (else 'foo) (#t 'bar)))
$1 = bar

The way this is done is by comparing with the original identifier given 
to syntax-rules. Technically, they are compared with free-identifier=? . 
This means that a use of the identifier matches the keyword iff both are 
unbound, or both are bound to the same lexical binding. However, this 
isn't the case here, as the keyword in the macro was unbound, but at the 
point of use, it has been bound by #:export.

Honestly, I think it is better to choose a different way of writing 
these macros that avoids this confusing issue. Try defining the 
operations at the same time as the enum so that the macro giving an enum 
member refers to the bindings of the operators. If you give more context 
on what you're trying to do, we could help more.

Best,
Jean

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