From Newsgroup: comp.lang.scheme

Zachary Beane wrote:

I am just trying to solve Ex 3.5 in Graham's ANSI Common Lisp book. I am reading it on my own and not as part of a university course. The task is:

define a function pos+, that takes a list as param and returns a list
that adds the position of each element of the list to the element's value. Thus:
(pos+ '(7 5 1 4))
returns:
(7 6 3 7)

...but it's also easy with DO:

(defun do-pos+ (orig-list)
(do ((i 0 (1+ i))
(list orig-list (cdr list))
(new-list nil (cons (+ (car list) i) new-list)))
((endp list) (nreverse new-list))))

(define (do-pos+ orig-list)
(do_ ((i 0 (+ 1 i))
(x :in orig-list)
(result :collect (+ i x)))
(#f @ result)))

(do-pos+ '(200 300 400))
===>
(200 301 402)

Given:

(define-syntax do_-aux
(syntax-rules ( <> @ :in :collect )
[ (do_-aux ((x what <>) more ...) (seen ...) stuff ...)
(do_-aux (more ...) (seen ... (x what what)) stuff ...) ]
[ (do_-aux ((x :in seq) more ...) seen (lets ...) (bool z ...) stuff ...)
(do_-aux ((x (and (pair? the-list) (car the-list)) <>) more ...)
seen
(lets ... (the-list seq))
((or (null? the-list) (begin (pop! the-list) #f) bool) z ...)
stuff ...) ]
[ (do_-aux ((accum :collect x) more ...) stuff ...)
(do_-aux ((accum '() (cons x accum)) more ...) stuff ...) ]
[ (do_-aux (spec more ...) (seen ...) stuff ...)
(do_-aux (more ...) (seen ... spec) stuff ...) ]
[ (do_-aux () seen lets (bool y ... @ result) stuff ...)
(do_-aux () seen lets (bool y ... (reverse result)) stuff ...) ]
[ (do_-aux () seen (lets ...) more ...)
(let (lets ...)
(do seen more ...))
] ))
(define-syntax do_
(syntax-rules ()
[ (do_ specs more ...)
(do_-aux specs () () more ...) ] ))
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