free-identifier =？之间的区别？和bound-identifier =？

Question

试图理解free-identifier =？和bound-identifier =？.任何人都可以给我使用free-identifier =的等效代码示例吗？将返回true并使用bound-identifier =？将返回false.

谢谢

Answer 1

这是一个例子:

(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)

在x介绍了go具有从它那一步扩张的标志,但x在compare-with-x没有,所以bound-identifier=?认为它们不同.

这是另一个例子:

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

这里go2还介绍了一个x带标记,而x给定go2的参数没有标记.相同的故事.