Hi, gustavo!
Thanks for your answer, it was very helpful.
So i rewrote my (substitute-term) macro like this:
(define-syntax (substitute-term stx)
(syntax-case stx ()
[(_ term-from term-to (body0 body ...))
#`#,(append (if (equal? (syntax-e #'body0) (syntax-e #'term-from))
(list #'term-to)
(list #'body0))
#'((substitute-term term-from term-to body) ...))]
[(_ term-from term-to body)
(if (equal? (syntax-e #'body)
(syntax-e #'term-from))
#'term-to
#'body)]))
The (syntax-e) parts is kept because (equal? #'body #'term-from) won't yield #t.
Now it works:
>(substitute-term and or (and #t #f))
#t
After a little bit tinkering I returned to (case) form, so it can substitute
not only single terms, but whole expressions:
(define-syntax (substitute-syntax stx)
(syntax-case stx ()
[(_ term-from term-to body)
(cond
[(equal? (syntax->datum #'body) (syntax->datum #'term-from)) #'term-to]
[(list? (syntax->datum #'body)) #`#,(append (if (equal? (syntax->datum
(car (syntax-e #'body)))
(syntax->datum
#'term-from))
(list #'term-to)
(list #`#,(car (syntax-e
#'body))))
(map (lambda (x)
(append (syntax-e
#'(substitute-syntax term-from term-to))
(list x)))
(cdr (syntax-e
#'body))))]
[else #'body])]))
>(substitute-syntax (and #t #f) or ((and #t #f) #f #t))
Expands into (or #t #f)
#t
And my final version of recursion looks like this:
(define-syntax recursion
(syntax-rules ()
[(_ (args ...) body ...)
((lambda (x) (x x))
(lambda (generated-label)
(lambda (args ...)
(substitute-syntax recursion (generated-label generated-label) body)
...)))]
[(_ label (args ...) body ...)
((lambda (x) (x x))
(lambda (label)
(lambda (args ...)
(substitute-syntax label (label label) body) ...)))]))
So it can be used without label, if there are no nested recursions:
>((recursion (n)
(if (zero? n)
1
(* n (recursion (sub1 n))))) 5)
5
Now it works, as intended.
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