Holger> ;; The following snippet behave differently whether Holger> ;; compiled or interpreted (i.e. evaluated at top-level) Holger> (setq gee 23) Holger> (format t "gee => ~d~%" gee) Holger> (defmacro fun () Holger> gee) Holger> (format t "(fun) => ~d~%" Holger> (fun)) Holger> (defun baz () Holger> (fun)) Holger> (format t "(baz) => ~d~%" Holger> (baz)) Holger> (setq gee 42) Holger> (format t "(setq gee 42)~%") Holger> (format t "(fun) => ~d~%" (fun)) Holger> (format t "(baz) => ~d~%" Holger> (baz)) Holger> Michael, in Scheme, to which values would baz and fun evaluate ? First, the code: (define gee 23) ; (1) (display "gee => ") (display gee) (newline) (define-syntax fun (syntax-rules () ((fun) gee))) ; this always refers to the binding established in (1) (display "(fun) => ") (display (fun)) (newline) (define (baz) ; note that the expansion of (fun) refers to the (fun)) ; of (1) (display "(baz) => ") (display (baz)) (newline) (set! gee 42) ; same binding, different value from now on (display "(set! gee 42)") (newline) (display "(fun) => ") (display (fun)) (newline) (display "(baz) => ") (display (baz)) (newline) Output: gee => 23 (fun) => 23 (baz) => 23 (set! gee 42) (fun) => 42 (baz) => 42 Note this is identical to the behavior of the code if you replace the definition of fun by (define (fun) gee) -- Cheers =8-} Chipsy Friede, Völkerverständigung und überhaupt blabla