Macros 语法中的语法引号、模板变量和省略号
我希望能够做的是转变Macros 语法中的语法引号、模板变量和省略号,macros,scheme,racket,Macros,Scheme,Racket,我希望能够做的是转变 (define count-suits (symbol-map-function hearts diamonds clubs spades)) 进入 我有lambda的尸体在处理 (define-syntax (generate-symbol-map stx) (syntax-case stx () ((gen...map enumerations ...) #'(make-hash (cons (quote enumerations) enumer
(define count-suits (symbol-map-function hearts diamonds clubs spades))
进入
我有lambda的尸体在处理
(define-syntax (generate-symbol-map stx)
(syntax-case stx ()
((gen...map enumerations ...)
#'(make-hash (cons (quote enumerations) enumerations) ...))))
但我有一个魔鬼的时间
(λ (#:hearts hearts
#:diamonds diamonds
#:clubs clubs
#:spades spades)
这就是我目前所拥有的
;; e.g. (weave '(1 3 5 7) '(2 4 6 8)) = '(1 2 3 4 5 6 7 8)
;; tested, works.
(define-for-syntax (weave list1 list2)
(cond ((empty? list1) list2)
((empty? list2) list1)
(else (list* (car list1)
(car list2)
(weave (cdr list1)
(cdr list2))))))
(define-syntax (symbol-map-function stx)
(syntax-case stx ()
((sym...ion symbols ...)
; What I'm trying to do here is splice the result of weaving the keywords,
; generated by format-id, with the symbols themselves, e.g. in the case of
; (symbol-map-function foo bar baz):
; #`(λ (#,@(weave '(#:foo #:bar #:baz) '(foo bar baz)))
; --> #`(λ (#,@'(#:foo foo #:bar bar #:baz baz))
; --> #`(λ (#:foo foo #:bar bar #:baz baz)
; I am using syntax-unquote-splicing because I am in syntax-quasiquote and the
; result of the expression is a list that I want to be spliced into the arguments.
#`(λ (#,@(weave (list (syntax-e (format-id #'symbols
"#:~a"
(syntax-e #'symbols))) ...)
(list #'(symbols ...))))
(generate-symbol-map symbols ...)))))
(列表(syntax-e(格式id:~a)(syntax-e#'符号))…)
旨在产生
(list (syntax-e (format-id #'foo "#:~a" (syntax-e #'foo)))
(syntax-e (format-id #'bar "#:~a" (syntax-e #'bar)))
(syntax-e (format-id #'baz "#:~a" (syntax-e #'baz))))
但有人告诉我在#符号之后缺少省略号。我尝试过以不同的方式处理代码,但没有任何真正的目的或见解,我也没有遇到任何有效的方法。
..
不能出现在模板之外,这意味着它们必须出现在#
符号之前的部分中。您可以编写#'(符号…
,但不能编写#符号…
在此之后,您可能需要使用,它将语法对象转换为语法对象列表
此外,您不能使用来生成关键字,因为格式id
将强制将结果作为符号,并将生成的id包含在管道中:
> (require racket/syntax)
> (format-id #'here "#:~a" 'auie)
#<syntax |#:auie|>
下面是符号映射函数的一个工作实现
:
(require (for-syntax racket/list))
(define-syntax (symbol-map-function stx)
(define (id->keyword id)
(datum->syntax id (string->keyword (symbol->string (syntax-e id)))))
(syntax-case stx ()
((_ id ...)
(andmap identifier? (syntax->list #'(id ...)))
(with-syntax ((lambda-list (append-map (lambda (id)
(list (id->keyword id) id))
(syntax->list #'(id ...)))))
#'(lambda lambda-list
(make-hash `((id . ,id) ...)))))))
我希望我知道一种比使用append map
更好的方法来组装lambda列表;欢迎改进。:-)
#lang racket
(require (for-syntax racket/syntax racket/list))
(define-syntax (foo stx)
(syntax-case stx ()
[(_ (sym ...) body ...)
(with-syntax ([kws (flatten
(map (λ(k)
(list
(string->keyword
(symbol->string
(syntax->datum k)))
k))
(syntax->list #'(sym ...))))]
)
#'(λ kws body ...))]))
; Test:
((foo (aa bb)
(list aa bb))
#:bb 'bbb
#:aa 'aaa)
; -> '(aaa bbb)
(require (for-syntax racket/list))
(define-syntax (symbol-map-function stx)
(define (id->keyword id)
(datum->syntax id (string->keyword (symbol->string (syntax-e id)))))
(syntax-case stx ()
((_ id ...)
(andmap identifier? (syntax->list #'(id ...)))
(with-syntax ((lambda-list (append-map (lambda (id)
(list (id->keyword id) id))
(syntax->list #'(id ...)))))
#'(lambda lambda-list
(make-hash `((id . ,id) ...)))))))