我很难理解Scheme的新宏系统。在路径的某个地方,我开始首先将“宏”作为函数编写,然后将其作为宏应用。
所以我的任务是改变以下结构:
;; highlight-rules: rule id, color and the regexp matches
(define highlight-rules
`((important ,(with-esc "[1;33m") ("foo"
"bob"))
(unimportant ,(with-esc "[1;30m") ("case of unimport"))
(urgent ,(with-esc "[1;31m") ("urgents"))))
进入这种cond
系列,匹配字符串编译为正则表达式:
;; just an example. `line` is an argument bound by the function application
(cond
((string-match (regexp ".*sudo:session.*") line)
(with-color *important* line))
(else line))
我写了一个似乎可以解决这个问题的函数:
;; (cdar highlight-rules) -> (colorstring list-of-rules)
(define (parse-highlight-rules rules)
;; aux function to do one 'class' of patterns
(define (class-of-rules colorstr rulelist)
(map (lambda (rule)
`((string-match ,(regexp rule)) (with-color ,colorstr line)))
rulelist))
(define (do-loop accumulator rules)
(let* ((highlight-group (cdar rules))
(colorstr (car highlight-group))
(grouprules (cadr highlight-group))
(acc* (append (class-of-rules colorstr grouprules) accumulator))
(rest (cdr rules)))
(if (null? rest)
acc*
(do-loop acc* rest))))
; wrap the list in cond.
`(apply cond ,(do-loop '() rules)))
使用给定的highlight-rules
函数返回正确的列表(除了应用apply
- 在clojure中,将使用拼接):
CSI> (parse-highlight-rules highlight-rules)
(apply cond (((string-match #<regexp>) (with-color "\x1b[1;31m" line))
((string-match #<regexp>) (with-color "\x1b[1;30m" line))
((string-match #<regexp>) (with-color #0="\x1b[1;33m" line))
((string-match #<regexp>) (with-color #0# line))))
但是如何处理呢?我已经坚持了一段时间。鸡计划是我的方言。
答案 0 :(得分:1)
将函数转换为宏的最简单方法是使用Chicken的explicit-renaming宏工具,它与Clojure的defmacro
类似(除了显式重命名宏需要一些额外的参数,可以是用于保护卫生。)
拼接的工作方式与Clojure中的拼接方式基本相同。语法为,@
。因此,以下应该有效:
(define-for-syntax (parse-highlight-rules rules)
;; ... insert missing code here ...
`(cond ,@(do-loop '() rules)))
(define-syntax highlight
(er-macro-transformer
(lambda (form rename compare)
(parse-highlight-rules (cdr form)))))