Emacs Lisp - mapcar 没有将函数应用于所有元素？

Question

我有一个函数，它接受一个列表并替换一些元素。我将它构建为一个闭包，这样自由变量就不能在函数外被修改。

(defun transform (elems)
  (lexical-let ( (elems elems) )
    (lambda (seq)
      (let (e)
    (while (setq e (car elems))
      (setf (nth e seq) e)
      (setq elems (cdr elems)))
    seq))))

我将其称为列表列表。

(defun tester (seq-list)
  (let ( (elems '(1 3 5)) )
    (mapcar (transform elems) seq-list)))

(tester (list (reverse (number-sequence 1 10)) 
          '("a" "b" "c" "d" "e" "f")))
=> ((10 1 8 3 6 5 4 3 2 1) ("a" "b" "c" "d" "e" "f"))

它似乎没有将函数应用于提供给 tester() 的列表的第二个元素。但是，如果我明确地将此函数应用于各个元素，它就可以工作...

(defun tester (seq-list)
  (let ( (elems '(1 3 5)) )
    (list (funcall (transform elems) (car seq-list))
          (funcall (transform elems) (cadr seq-list)))))

(tester (list (reverse (number-sequence 1 10)) 
          '("a" "b" "c" "d" "e" "f")))
=> ((10 1 8 3 6 5 4 3 2 1) ("a" 1 "c" 3 "e" 5))

如果我使用与上述相同的概念编写一个简单的函数，mapcar 似乎可以工作...我可能做错了什么？

(defun transform (x)
  (lexical-let ( (x x) )
    (lambda (y) 
      (+ x y))))

(defun tester (seq)
  (let ( (x 1) )
    (mapcar (transform x) seq)))

(tester (list 1 3))    
=> (2 4)

谢谢

Answer 1

闭包变量 elems 在第一次调用闭包后被设置为 nil；所以所有后续调用只会看到零。您的第二个示例有效，因为 transform 的每个实例都会产生一个新的闭包。

这应该有效:

(defun transform (elems)
  (lexical-let ((elems elems))
    (lambda (seq)
      (dolist (e elems seq)
        (setf (nth e seq) e)))))