I see. Okay, thanks for clarifying. On Sat, Jul 16, 2016 at 8:05 PM, Alex Knauth <alexan...@knauth.org> wrote: > >> On Jul 16, 2016, at 7:18 PM, David Storrs <david.sto...@gmail.com> wrote: >> >> Wow, that is a lot of problems. Thanks for taking the time to >> comment; this was in large part an effort to learn macros, so it's >> helpful to get this kind of feedback. > >> On Sat, Jul 16, 2016 at 2:48 PM, Alex Knauth <alexan...@knauth.org> wrote: > >>> A few problems with this. >>> >>> One, you shouldn't put extra (lambda () ...)s around the #:when conditions. >>> A lambda is a truthy value, since it's not false, so those #:when checks >>> weren't checking anything at all. >> >> I thought the #:when ran the predicate that it was given and reacted >> based on the truth value of the response? > > The #:when takes a boolean. It works like the #:when clause in match, if > you've used that. Or like the condition of an if expression where the > then-branch keeps going in the same clause and the else-branch makes it go to > the next clause. > >> Would (string? pred) have worked there instead? > > I'm not sure what you mean. > > #lang racket > ;; no, I don't want a for-syntax, I'm using it at run-time, as a match-like > form > (require syntax/parse) > (syntax-parse (syntax ignored) > [_ > #:when (string? "hello") > 'string]) > > This tests whether "hello" is a string. But for a pattern-variable like pred > it almost can work like that, but not quite. > > (syntax-parse (syntax (foo "hello")) > [(foo a) > #:when (string? a) > 'string]) > > This gives the error `a: pattern variable cannot be used outside of a > template`. A template means something like the syntax form. However even with > that fixed this doesn't work as expected: > > (syntax-parse (syntax (foo "hello")) > [(foo a) > #:when (string? (syntax a)) ; writing (string? #'a) would mean the same > thing > 'string]) > > The string? predicate returns false, so this fails as well. The problem is > that (syntax a) returns a syntax object, but you want the contents inside the > syntax object, because that's where the string (or otherwise) will be. To do > that you can use syntax-e (one-level-down) or syntax->datum (pure > s-expression with no syntax-object sub-expressions). > > (syntax-parse (syntax (foo "hello")) > [(foo a) > #:when (string? (syntax-e (syntax a))) > 'string]) > > This succeeds, producing 'string. Using syntax->datum will also work here: > > (syntax-parse (syntax (foo "hello")) > [(foo a) > #:when (string? (syntax->datum (syntax a))) > 'string]) > >>> Two, once you remove those lambdas, (procedure? #'pred) will always fail >>> because #'pred isn't the actual predicate, it's an expression, which might >>> eventually evaluate to predicate, or it might evaluate something else. At >>> compile-time, you just don't know yet. What you could do is check that it's >>> an identifier, with (identifier? #'pred). >> >> Would (procedure? (syntax->datum #'pred)) have worked there? pred >> either is or is not a procedure, and that should be known at compile >> time. > > For simple literal data like strings, (string? (syntax->datum #'pred)) will > work, since literal data is known at compile-time. > > However procedures, although they are values, they only become values at > run-time. At compile-time, this pred is only an expression, so (procedure? > (syntax->datum #'pred)) will return false. > > A procedure given to you by a user in something like > (throws (boom) exn:fail? "should trigger the proc form") > Will not be known at compile time. (syntax pred) is an identifier, and > (syntax->datum (syntax pred)) will return a symbol, not a procedure. > > At compile-time that's all you know. It could be an identifier that happens > to be bound to a procedure, or it could be an identifier bound to a number, > or it could have no definition at all. > > So, for simple literal data like strings, it is known at compile-time, but > for more complicated things like procedures, it usually isn't. > > Alex Knauth >
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