lisp - 在 Lisp 中一次定义 n 个函数

Question

假设我想做以下事情:

(loop for i from 1 to n do
    (defun ith(lst)
        (nth i lst)))

显然我真正想做的是:

(defun 1th(lst)(nth 1 lst))
(defun 2th(lst)(nth 2 lst))
(defun 3th(lst)(nth 3 lst))
(defun 4th(lst)(nth 4 lst))
......
(defun 100th(lst)(nth 100 lst))

我该怎么做？

Answer 1

给你。请注意，|1th| 将返回第二个值:

(defmacro make-nths (n)
  `(progn
     ,@(loop for i from 1 to n
             collecting `(defun ,(intern (format nil "~ath" i)) (list)
                           (nth ,i list)))))

正如 Xach 在评论中指出的那样，macrolet 在这里可能更可取，因为您实际上并不需要全局定义的宏:

(macrolet ((make-nths (n)
             `(progn
                ,@(loop for i from 1 to n
                        collect `(defun ,(intern (format nil "~ath" i)) (list)
                                   (nth ,i list))))))
  (make-nths 3)) ; choose n here

最后，这是 ninjaaa 的非宏解决方案的工作版本:

(loop for i from 1 to 3 ; choose n here
      do (let ((i i))   ; introduce new bindings for each iteration
           (setf (symbol-function (intern (format nil "~ath" i)))
                 (lambda (list) (nth i list)))))