Re: define-module, #:export and export

[Jean Abou Samra]

Le 08/01/2023 à 16:18, yarl baudig a écrit :
> Obviously...
>
> Ok, I tried more.
> Apparently,
>
> (define-syntax define-enumerate-type
>    (syntax-rules ()
>      ((_ name->int (name id) ...)
>       (define-syntax name->int
>         (syntax-rules (name ...)
>          ((_ name) id) ...)))))
>
> works the same as
>
> (define-syntax define-enumerate-type
>    (syntax-rules ()
>      ((_ name->int (name id) ...)
>       (define-syntax name->int
>         (lambda (x)
>          (syntax-case x ()
>            ((_ n) (free-identifier=? #'name #'n) #'id) ...))))))
>
> . Now what about
>
> (define-syntax define-enumerate-type
>    (syntax-rules ()
>      ((_ name->int (name id) ...)
>       (define-syntax name->int
>         (lambda (x)
>          (syntax-case x ()
>            ((_ n) (equal?
>                    (syntax->datum #'name)
>                    (syntax->datum #'n))
>             #'id) ...))))))
>
> ?



Yes, that was Maxime's suggestion (c). It essentially means you take the 
symbol
value of the argument to the macro without caring about whether it is bound
or renamed.

If you're doing this, maybe it's simpler to just turn the macro into a 
function
that accepts symbol values and call it as (my-enum 'foo) instead of 
(my-enum foo)?

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