Question

假设我想要执行以下操作：

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

显然我真正想做的是以下内容：

(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

你走了。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可能更适用于此，因为您并不真正需要全局定义的宏：

(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非宏观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)))))

Answer 2

试试这个：

(loop for i from 1 to n do
   (setf (symbol-function (intern (format nil "~dth" i)))
     #'(lambda (lst) (nth i lst))))

在Lisp中一次定义n个函数

2 个答案: