difference between free-identifier=? and bound-identifier=?

Question

Trying to understand free-identifier=? and bound-identifier=?. Can anyone give me equivalent code examples where using free-identifier=? would return true and using bound-identifier=? would return false.

Thanks

Answer 1

Here's an example:

(define-syntax (compare-with-x stx)
  (syntax-case stx ()
    [(_ x-in)
     (with-syntax ([free=? (free-identifier=? #'x-in #'x)]
                   [bound=? (bound-identifier=? #'x-in #'x)])
       #'(list free=? bound=?))]))

(define-syntax go
  (syntax-rules ()
    [(go) (compare-with-x x)]))

(go) ;; => '(#t #f)

The x introduced by go has a mark from that expansion step on it, but the x in compare-with-x doesn't, so bound-identifier=? considers them different.

Here's another example:

(define-syntax (compare-xs stx)
  (syntax-case stx ()
    [(_ x1 x2)
     (with-syntax ([free=? (free-identifier=? #'x1 #'x2)]
                   [bound=? (bound-identifier=? #'x1 #'x2)])
       #'(list free=? bound=?))]))

(define-syntax go2
  (syntax-rules ()
    [(go2 x-in) (compare-xs x-in x)]))

(go2 x) ;; => '(#t #f)

Here go2 also introduces an x with a mark, whereas the x given to go2 as an argument does not have a mark. Same story.