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chez-openbsd/ta6ob/examples/setof.ss
2022-08-09 23:28:25 +02:00

53 lines
2.1 KiB
Scheme

;;; setof.ss
;;; Copyright (C) 1996 R. Kent Dybvig
;;; from "The Scheme Programming Language, 2ed" by R. Kent Dybvig
;;; Permission is hereby granted, free of charge, to any person obtaining a
;;; copy of this software and associated documentation files (the "Software"),
;;; to deal in the Software without restriction, including without limitation
;;; the rights to use, copy, modify, merge, publish, distribute, sublicense,
;;; and/or sell copies of the Software, and to permit persons to whom the
;;; Software is furnished to do so, subject to the following conditions:
;;;
;;; The above copyright notice and this permission notice shall be included in
;;; all copies or substantial portions of the Software.
;;;
;;; THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
;;; IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
;;; FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL
;;; THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
;;; LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING
;;; FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
;;; DEALINGS IN THE SOFTWARE.
;;; set-of uses helper syntactic extension set-of-help, passing it
;;; an initial base expression of '()
(define-syntax set-of
(syntax-rules ()
((_ e m ...)
(set-of-help e '() m ...))))
;;; set-of-help recognizes in, is, and predicate expressions and
;;; changes them into nested named let, let, and if expressions.
(define-syntax set-of-help
(syntax-rules (in is)
((_ e base)
(set-cons e base))
((_ e base (x in s) m ...)
(let loop ((set s))
(if (null? set)
base
(let ((x (car set)))
(set-of-help e (loop (cdr set)) m ...)))))
((_ e base (x is y) m ...)
(let ((x y)) (set-of-help e base m ...)))
((_ e base p m ...)
(if p (set-of-help e base m ...) base))))
;;; set-cons returns the original set y if x is already in y.
(define set-cons
(lambda (x y)
(if (memv x y)
y
(cons x y))))