difference between free-identifier=? and bound-identifier=?
Trying to understand free-identifier=? and bound-identifier=?. Can anyone give me equivalent code examples wher开发者_C百科e using free-identifier=? would return true and using bound-identifier=? would return false.
Thanks
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.
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