我的 lisp 解释器用 JavaScript 编写的宏有问题。问题出在这段代码中：
(define log (. console "log"))
(define (alist->object alist)
"(alist->object alist)
Function convert alist pairs to JavaScript object."
(if (pair? alist)
((. alist "toObject"))))
(define (klist->alist klist)
"(klist->alist klist)
Function convert klist in form (:foo 10 :bar 20) into alist
in form ((foo . 10) (bar . 20))."
(let iter ((klist klist) (result '()))
(if (null? klist)
result
(if (and (pair? klist) (pair? (cdr klist)) (key? (car klist)))
(begin
(log ":::" (cadr klist))
(log "data" (. (cadr klist) "data"))
(iter (cddr klist) (cons (cons (key->string (car klist)) (cadr klist)) result)))))))
(define (make-empty-object)
(alist->object '()))
(define empty-object (make-empty-object))
(define klist->object (pipe klist->alist alist->object))
;; main function that give problems
(define (make-tags expr)
(log "make-tags" expr)
`(h ,(key->string (car expr))
,(klist->object (cadr expr))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,@(map make-tags (cdaddr expr))))
(caddr expr)))))
(define-macro (with-tags expr)
(make-tags expr))
我使用以下代码调用此宏：
(define (view state actions)
(with-tags (:div ()
(list (:h1 () (value (cdr (assoc 'count (. state "counter")))))
(:button (:onclick (lambda () (--> actions (down 1)))) "-")
(:button (:onclick (lambda () (--> actions (up 1)))) "+")))))
这应该扩展到几乎相同的代码：
(define (view state actions)
(h "div" (make-empty-object)
(list->array (list
(h "h1" (make-empty-object) (value (cdr (assoc 'count (. state "counter")))))
(h "button" (klist->object `(:onclick ,(lambda () (--> actions (down 1))))) "-")
(h "button" (klist->object `(:onclick ,(lambda () (--> actions (up 1))))) "+")))))
此功能有效。我使用调用主函数的宏的扩展代码有问题，不知道 LIPS 在找到时应该如何表现：
(:onclick (lambda () (--> actions (down 1))))
内部代码，您尝试像这样处理它：
,(klist->object (cadr expr))
现在我的 lisp 工作，将 lambda 标记为数据（将数据标志设置为 true，这是防止从宏对某些代码进行递归评估的技巧）并且klist->object函数将 lambda 代码作为列表，而不是函数。
这应该如何在 Scheme 或 Common Lisp 中工作？应该klist->object获取函数对象（lambda 被评估）还是以 lambda 作为第一个符号的列表结构？如果是第二个，那么我应该如何编写函数和宏来评估 lambda，我应该使用 eval （对我来说有点骇客）。
抱歉不知道如何测试这个，有更多无错误的 LISP。
编辑：
我试图在诡计中应用来自@jkiiski 的提示（因为在我的 lisp 中它不起作用）
;; -*- sheme -*-
(define nil '())
(define (key? symbol)
"(key? symbol)
Function check if symbol is key symbol, have colon as first character."
(and (symbol? symbol) (eq? ":" (substring (symbol->string symbol) 0 1))))
(define (key->string symbol)
"(key->string symbol)
If symbol is key it convert that to string - remove colon."
(if (key? symbol)
(substring (symbol->string symbol) 1)))
(define (pair-map fn seq-list)
"(seq-map fn list)
Function call fn argument for pairs in a list and return combined list with
values returned from function fn. It work like the map but take two items from list"
(let iter ((seq-list seq-list) (result '()))
(if (null? seq-list)
result
(if (and (pair? seq-list) (pair? (cdr seq-list)))
(let* ((first (car seq-list))
(second (cadr seq-list))
(value (fn first second)))
(if (null? value)
(iter (cddr seq-list) result)
(iter (cddr seq-list) (cons value result))))))))
(define (klist->alist klist)
"(klist->alist klist)
Function convert klist in form (:foo 10 :bar 20) into alist
in form ((foo . 10) (bar . 20))."
(pair-map (lambda (first second)
(if (key? first)
(cons (key->string first) second))) klist))
(define (h props . rest)
(display props)
(display rest)
(cons (cons 'props props) (cons (cons 'rest rest) nil)))
(define (make-tags expr)
`(h ,(key->string (car expr))
(klist->alist (list ,@(cadr expr)))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,@(map make-tags (cdaddr expr))))
(caddr expr)))))
(define-macro (with-tags expr)
(make-tags expr))
(define state '((count . 10)))
(define xxx (with-tags (:div ()
(list (:h1 () (cdr (assoc 'count state)))
(:button (:onclick (lambda () (display "down"))) "-")
(:button (:onclick (lambda () (display "up"))) "+")))))但出现错误：
错误：未绑定的变量：：onclick
我找到了我的 lisp 的解决方案，这是代码：
(define (pair-map fn seq-list)
"(seq-map fn list)
Function call fn argument for pairs in a list and return combined list with
values returned from function fn. It work like the map but take two items from list"
(let iter ((seq-list seq-list) (result '()))
(if (null? seq-list)
result
(if (and (pair? seq-list) (pair? (cdr seq-list)))
(let* ((first (car seq-list))
(second (cadr seq-list))
(value (fn first second)))
(if (null? value)
(iter (cddr seq-list) result)
(iter (cddr seq-list) (cons value result))))))))
(define (make-tags expr)
(log "make-tags" expr)
`(h ,(key->string (car expr))
(alist->object (quasiquote
;; create alist with unquote for values and keys as strings
,@(pair-map (lambda (car cdr)
(cons (cons (key->string car) (list 'unquote cdr))))
(cadr expr))))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,@(map make-tags (cdaddr expr))))
(caddr expr)))))
因此，在我的代码中，我正在编写某种元宏，我正在编写 quasiquote 作为列表，它将得到评估，就像我在原始代码中使用的一样：
(klist->object `(:onclick ,(lambda () (--> actions (down 1)))))
我正在使用alist->object新函数pair-map，所以我可以取消引用值并将键符号转换为字符串。
这是应该如何在方案中实施的吗？不确定我是否需要修复我的 lisp 或宏在那里正常工作。