Dear list members,
In the Racket Guide 16.1.7 Extended Example: Call-by-Reference Functions
it shows how with macros to set up call by reference versions of
Racket functions.
I'm a little bit confused on why for the example, the macro (swap x y)
was used as the example function to turn into a call-by-reference
function, since doesn't the macro (swap x y) already act like a
call-by-reference function.
Ie. i the example I can't see any difference between the effects of:
(swap x y)
and the effects of:
(define-cbr (swap-by-reference x y) (swap x y))
(swap-by-reference x y)
In this particular case is there some difference in the effects of
(swap x y) and (swap-by-reference x y) ?
Here is the full code of the example as I've entered it into DrRacket
#lang racket
(define-syntax-rule (swap x y)
(let ([tmp x])
(set! x y)
(set! y tmp)))
(define-syntax-rule (define-get/put-id id get put!)
(define-syntax id
(syntax-id-rules (set!)
[(set! id e) (put! e)]
[(id a (... ...)) ((get) a (... ...))]
[id (get)])))
(define-syntax f
(syntax-rules ()
[(id actual ...)
(do-f (lambda () actual)
...
(lambda (v)
(set! actual v))
...)]))
(define-syntax-rule (define-cbr (id arg ...) body)
(begin
(define-syntax id
(syntax-rules ()
[(id actual (... ...))
(do-f (lambda () actual)
(... ...)
(lambda (v)
(set! actual v))
(... ...))]))
(define-for-cbr do-f (arg ...)
() ; explained below...
body)))
(define-syntax define-for-cbr
(syntax-rules ()
[(define-for-cbr do-f (id0 id ...)
(gens ...) body)
(define-for-cbr do-f (id ...)
(gens ... (id0 get put)) body)]
[(define-for-cbr do-f ()
((id get put) ...) body)
(define (do-f get ... put ...)
(define-get/put-id id get put) ...
body)]))
(define-cbr (swap-by-ref a b)
(swap a b))
(let ([x 1] [y 2])
(swap-by-ref x y)
(list x y))
(let ([x 1] [y 2])
(swap x y)
(list x y))
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