Is SBCL retarded?

or am i just doing something wrong?


(defmacro test (&body c)
  `(progn ,(test2 c)))

(defmacro test2 (&body c)
  (format t "~A~%" c)
  `(progn (car ,c) ,@c))

test2 compiles just fine
upon trying to compile test i get:

; in: DEFMACRO TEST
;     COMMON-LISP::C
;
; caught STYLE-WARNING:
;   undefined function: C
;
; compilation unit finished
;   Undefined function:
;     C
;   caught 1 STYLE-WARNING condition


How the fuck does it see C as a function when it's clearly in the argument position for test2?

>>1
Because C in `(progn ,(test2 c))) isn't bound to outer's C value, but passed as symbol C.

think about what a macro does.

- It takes input
- It turns that input into S-expressions
- Returns the output
- Compiler then evals the output

test is basically doing:

(eval (progn (eval (progn (car c) ,@c))))
=>(eval (progn ,@c))
=>,@c

and in the manner you're calling test2,
c = ( (exps...) )

thus the final result would be interpreting the first expression in c as a function.



tl;dr: test2 should be a defun, not defmacro

>>2
i.e. macros never evaluate their params - that is whole point of them. OP should define test2 as (defun test2 (&body c)

I order to say what you are doing wrong, we need to know what you are trying to do.

>>4
>>3

(defun test2 (&body c)
  (format t "~A~%" c)
  `(progn (car ,c) ,@c))
now errors:


; caught ERROR:
;   Bad lambda list keyword &BODY in: (&BODY C)
;
; compilation unit finished
;   caught 1 ERROR condition

>>6
use &rest instead of &body

OP all you need is:

(defmacro test (&body c)
  `(progn (test2 ,@c)))

SBCL will automatically expand test2 when compiling, you don't need to invoke it manually like that