On 04/10/2011 12:31 AM, Veer wrote:
Hello,
I am unable to understand how free-identifier=? and bound-identifier=? works?
For example suppose I have a code :
(lambda (x y) (let ([x 2]) x))
then how do I determine if last x in the body of let is not bound by first
parameter to lambda? , using either free-identifier=? or bound-identifier=? .
You can only get answers to questions like this *after* expanding the
program. You might have a notion of the scoping rules of lambda and let,
but free-identifier=? et al don't; they can only tell you what the macro
expander has discovered about the program's binding structure.
Before any expansion, the macro expander hasn't made any discoveries, so
all xs look alike.
After expansion, you get something that looks like this:
(lambda (x y) (let-values ([(x) '2]) x)
so you'll have to rewrite your pattern, but now you can ask questions
using free-identifier=?. (IIRC, free-identifier=? and bound-identifier=?
are equivalent on fully-expanded programs.)
During macro expansion, free-identifier=? and bound-identifier=? compare
identifiers based on the discoveries available at that point in time.
(free-identifier=? x y) means if these two identifiers were turned into
references right now, would they refer to the same binding. It's how
cond, for example, recognizes the else keyword: does the identifier the
macro gets refer to the same binding as a known good reference to else.
(bound-identifier=? x y) means if one of the identifiers were turned
into a binding occurrence and the other were turned into a reference in
its scope, eg (lambda (x) y), would the second refer to the first. This
is usually only used when you're implementing a new binding form to
check for duplicates, etc, or for really odd macros that act kind of
like binding forms but don't use Racket's core binding forms to do it.
Ryan
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