[racket] How does free-identifier=? and bound-identifier=? works?

Veer wrote:
> I  am  unable  to  understand  how  free-identifier=?  and
> bound-identifier=? works?
[...]
> When I use them ,they both produces #t .

You are misunderstanding the  meaning of the function names;
you may try  to read [1] which is  an annotated section from
the R6RS standard.

  However,  your  true  misunderstanding  is to  think  that
identifiers in the input form  of a macro transformer have a
"meaning", for  example LET is  a binding syntax;  they have
not: identifiers in an input form are just identifiers, they
do  not affect the  binding of  other identifiers.   In your
example:

  (check (lambda (x y) (let ([x 2]) x)))

the identifiers LAMBDA and LET  in the input form of the use
of CHECK are "just identifiers",  they do not influence X in
any way; only if you put  into the output form of CHECK they
may behave  like the syntaxes  of the language (if  they are
in the correct position).

  Also  notice that your  CHECK macro  (as I  understand it)
does not do what  you want to do; to do what  you want to do
you should (notice the use of #` and #,):

#!r6rs
(import (rnrs))

(define-syntax check
  (lambda (stx)
    (syntax-case stx (lambda let)
      [(_ (lambda (x y ...) (let ([a b]) c)))
       #`(values #,(free-identifier=? #'x #'c)
                 #,(bound-identifier=? #'x #'c))])))

(let-values (((r1 r2)
	      (check (lambda (x y) (let ([x 2]) x)))))
  (write (list r1 r2))
  (newline))

HTH

[1] <http://marcomaggi.github.com/docs/nausicaa.html/stdlib-syntax_002dcase-identifier.html>

P.S.  It takes time understand this stuff; also, some people
think that  FREE-IDENTIFIER=? and BOUND-IDENTIFIER=?  are an
unfortunate choice of names, which induces misunderstanding.
-- 
Marco Maggi