The `=defun` macro introduces the `*cont*` binding as the argument of `=foo`. Hygienic macro expansion ensures that this macro-introduced bindings does not capture uses of `*cont*` that appear in the macro use. Specifically, the `*cont*` in
(=defun (foo x) *cont*) refers to the module-level definition, and not to the macro-introduced argument. To implement a macro along the lines of `=defun`, probably you want to use `define-syntax-parameter` and `syntax-parameterize`. At Fri, 7 Feb 2014 08:37:04 -0700, Kevin Forchione wrote: > I’m trying to resolve why I get differing results in a let form when using a > macro versus using the expanded function version. Here’s an example using > racket 5.93: > > #lang racket > > (require (for-syntax racket/syntax)) > > (define *cont* identity) > > (define-syntax (=defun stx) > (syntax-case stx () > [(_ (name parm ...) body ...) > (with-syntax ([f (format-id stx "=~a" #'name)] > [*cont* (format-id stx "~a" '*cont*)]) > #'(begin > (define-syntax-rule (name parm ...) > (f *cont* parm ...)) > (define (f *cont* parm ...) body ...)))])) > > (=defun (foo x) *cont*) > > (let ([*cont* cdr]) > (foo 10)) > => #<procedure:identity> > > (let ([*cont* cdr]) > (=foo *cont* 10)) > => #<procedure:cdr> > > Why does the macro return identity, while the function returns cdd? > > Thanks! > ____________________ > Racket Users list: > http://lists.racket-lang.org/users ____________________ Racket Users list: http://lists.racket-lang.org/users
