"Apparent identifier binding" just shows the result of the
identifier-binding function. It is labeled "apparent" because macro
expansion is a discovery process, and if the identifier is in an
unexpanded part of the program, the macro expander might discover
additional binding forms that change its meaning before getting to the
identifier itself.
The binding an identifier refers to depends on the marks, but it also
depends on the environment where the identifier first occurred.
Here's one example:
(define-syntax-rule (m)
x)
(define-syntax (m2 stx)
(syntax-case stx ()
[(m2)
(syntax-local-introduce #'x)]))
(lambda (x)
(m)
(m2)
x)
In the lambda body, the x's produced by m and m2 both have apparent
identifier binding of "none", but the direct use of x has binding
"lexical". The x produced by m has a mark from the expansion of m, but
the one produced by m2 has no mark, because m2 cancels it out using
syntax-local-introduce.
Here's another example:
(define-syntax (m3 stx)
(syntax-case stx ()
[(m3)
(with-syntax ([x1 #'x]
[x2 (let ([x 'whatever]) #'x)])
#'(void x1 x2))]))
(m3)
The first x produced by m3 has a binding of "none", but the second one
has a binding of "lexical" due to the let binding *in the implementation
of the macro*. If the expander gets to the second x reference, it will
raise an "identifier used out of context" error; that error is usually
caused by a macro like the one above using the same name as an
implementation-level binding and in the generated syntax.
Ryan
On 01/15/2015 01:28 PM, Spencer Florence wrote:
Hey All,
I'm trying to understand why two identifiers I have don't reference the
same value. If I look at them in the macro stepper all of their
properties and marks are the same, with the exception of "Apparent
identifier binding". One has "lexical (all phases)" and the other
"none". What is "Apparent identifier binding" and how does it get
determined?
--spencer
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