To give a simple example:
(define-macro-variable _iota 0) ; define-macro-variable does not really exist
(define-syntax (iota stx)
(syntax-case stx ()
((iota)
(let ((i _iota))
(set! _iota (+ i 1))
#`#,i))))
Such that given:
(define zero (iota))
(define one-two-three (list (iota) (iota) (iota)))
(define (four) (iota))
the following should all evaluate to #t
:
(equal? zero 0)
(equal? one-two-three '(1 2 3)) ; possibly in a different order
(equal? (four) 4)
(equal? (four) 4)
(equal? (four) 4)
Is there any real racket feature that does what define-macro-variable
is supposed to do in the example above?
EDIT:
I found a work-around:
(define-syntaxes (macro-names ...)
(let (macro-vars-and-vals ...)
(values macro-bodies-that-nead-the-macro-vars ...)))
but I would prefer a solution that does not require all the macros that use the macro variables to be in one expression.
You want define-for-syntax
(in Racket).
(define-for-syntax _iota 0)
(define-syntax (iota stx)
(syntax-case stx ()
((iota)
(let ((i _iota))
(set! _iota (+ i 1))
#`#,i))))
(define zero (iota))
(define one-two-three (list (iota) (iota) (iota)))
(define (four) (iota))
(equal? zero 0)
(equal? one-two-three '(1 2 3))
(equal? (four) 4)
(equal? (four) 4)
(equal? (four) 4)
produces all true.