Based on what you have written so far, the `versions` macro has no
sub-expressions, so you shouldn't use `expr/c` at all. It requires version
bounds to be in the form of literal strings. So you could describe the
macro using a grammar as follows:
Expression ::= .... | (versions Version ...)
Version ::= VersionBound | (VersionBound VersionBound)
VersionBound ::= String
I think what you want to do is refine VersionBound so that it only accepts
strings of a certain form. The best way to do that is with a separate
syntax class that matches a string and then puts additional side-conditions
on it (using `#:when`, etc). That is, you check the `valid-version?`
predicate at compile-time.
By the way, you should also avoid treating the literal strings that your
macro receives as if they were also expressions. A syntax object containing
a literal string is *not necessarily* a string-valued expression. Once your
macro views and validates something as a literal string, the proper way to
convert it to a run-time expression is to explicitly quote it. Consider the
following test case:
(let-syntax ([#%datum (lambda (stx) #'(exit '0))]) (versions ("7.0"
"7.7.0.5") "6.5"))
If your macro produces eg (list (make-version-range (quote "7.0") (quote
"7.7.0.5")) (quote "6.5")), then it's fine; if it produces (list
(make-version-range "7.0" "7.7.0.5") "6.5"), then it would exit. This
particular example is unlikely to happen in practice, but I think it is
useful to think clearly about how interpret each argument of a macro. Treat
it as a literal string or as an expression, but not both.
A different design would be to say that VersionBound is an expression that
produces a string. That would cause problems with your current grammar,
because you couldn't tell whether `(f "1.2.3")` was a single version (whose
value is produced by a function call) or by a range (whose lower bound is
the variable f). But you could change the grammar to avoid that problem.
Then you could use `expr/c` to wrap the expressions to check that at run
time they produced strings of the proper form.
Ryan
On Thu, Dec 17, 2020 at 12:55 AM Sage Gerard <[email protected]> wrote:
> 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 <[email protected]>
> 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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