If you change datum->syntax's first argument to #'argument, then it does what you want.
In general if you want the identifiers created by datum->syntax to be visible to the user, you need to supply a piece of syntax that you got from the user. That's why calling "long" directly works but not if it's called via "short". On Mon, Sep 2, 2013 at 8:08 PM, Tim Jervis <[email protected]> wrote: > Hello, > > I have been trying to create a macro to make some arbitrary definitions, > with success. The following defines a structure (posn) and a value (val): > > (define-syntax (long syntax-object) > (syntax-case syntax-object () > [(_ argument) > (let ([make-id (lambda (x) (datum->syntax syntax-object x))]) > (with-syntax ([posn (make-id 'posn)] > [val-s (make-id 'val)]) > #'(begin (printf "\tfrom the \"long\" macro, defining a structure > and a value\n") > (struct posn (x y)) > (define val-s 12))))])) > > > So: > > (long argument) > > > defines posn and val. > > However, if I happen to use this macro indirectly, the definitions aren't > visible. For example, if I define: > > (define-syntax (short syntax-object) > (syntax-case syntax-object () > [(_ argument) > #'(long argument)])) > > > and then call > > (short argument) > > > posn and val are not defined, even though the text in the printf buried in > the first macro does appear. > > I tried this in Dr Racket. > > Can anyone tell me where I'm going wrong? > > Kind regards, > > > > Tim > > > ____________________ > Racket Users list: > http://lists.racket-lang.org/users > ____________________ Racket Users list: http://lists.racket-lang.org/users
