This is not the same as a recursive macro.
Define this:
(define-syntax f
(syntax-rules ()
((f) f)))Now what is the result of (f)? There are three possible results:
When syntax-rules is implemented on top of syntax-case, the result
is likely to be (3).
| system | result |
|---|---|
| Bigloo | 1 ("f unbound") |
| Biwa | 2 ("a macro") |
| Chibi | 3 ("syntax error") |
| Chicken | 1 ("f unbound") |
| Chez | 3 ("syntax error") |
| Cyclone | 2 ("a macro") |
| Gambit | 3 ("syntax error") |
| Gauche | 2 ("a macro") |
| Guile | 3 ("syntax error") |
| Kawa | 2 ("a macro") |
| LIPS | 2 ("a macro") |
| Loko | 3 ("syntax error") |
| MIT | 3 ("syntax error") |
| Peroxide | 3 ("syntax error") |
| Sagittarius | 3 ("syntax error") |
| Scheme 48 | 1 ("f unbound") |
| Scheme 9 | see below |
| STklos | 1 ("f unbound") |
| Unsyntax | 3 ("syntax error") |
For Scheme 9 From Empty Space, we get:
> (define-syntax f
(syntax-rules ()
((f) f)))
> (f)
*** error: undefined symbol: _
*** trace: alpha-conv form conv subst expandIt isn't useful to compare this to Common Lisp and Emacs Lisp (because they are Lisp-2, and there is no way to reference the macro itself in its expansion). But we can do that in Clojure:
Clojure 1.10.2
user=> (defmacro f [] f)
#'user/f
user=> (f)
#object[user$f 0x3ff57625 "user$f@3ff57625"]It seems that it does return the macro, but checking its type doesn't tell much:
user=> (type (f))
user$f