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

From: Ryan Culpepper (ryanc at ccs.neu.edu)
Date: Sun Apr 10 07:29:56 EDT 2011

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.


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