Typos:
- "*" remove a bound ==> "*" removes a bound
- All examples should read (versions ...), not (version ...)
~slg
‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐
On Wednesday, December 16, 2020 6:27 PM, Sage Gerard <s...@sagegerard.com>
wrote:
> I'm trying to learn how to write syntax classes. My intended macro expresses
> a set of Racket versions, either as inclusive intervals or as exact versions.
> In an interval, "*" remove a bound.
>
> - (version "6.5") means exactly version "6.5", as does (version ("6.5" "6.5"))
>
> - (versions ("7.0" "7.7.0.5")) means the inclusive interval between version
> 7.0 and 7.7.0.5
> - (versions ("7.0" "7.7.0.5") "6.5"): union of the above two items
> - (versions ("6.0" "*")): all Racket versions >= 6.0
> - (versions "*"), (versions ("*" "*")): all Racket versions
>
> I was able to define the syntax class without much issue:
>
> (define-syntax-class racket-version-selection
> #:attributes (min max)
> (pattern (min:string max:string))
> (pattern (~and (~var v string)
> (~bind [min #'v]
> [max #'v]))))
>
> Now I want each attribute-bound expression V to satisfy (or (valid-version?
> V) (equal? V "*")). Where I'm stuck is how I can use #:declare with (expr/c)
> here. From what I understand, expr/c does not really mean much because it
> accepts an expression (as in the expr syntax-class), not attributes.
>
> The only way I can think to fix this is to perform an additional syntax-parse
> so that I can use the attributes in an expression for expr/c to consume. But
> is it possible to do everything I'm thinking of in just one syntax class?
>
> ~slg
>
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