common-lisp - 将字符串转换为函数

Question

我有字符串“+”，我想将其转换为#'+。基本上我想将字符串映射到相应的函数即“myfunname”到#'myfunname。

Answer 1

这几乎肯定是一个 XY 问题:为什么要从字符串开始？

如果您确实需要将字符串转换为函数，那么类似下面的代码就可以做到这一点，并且比使用read安全得多:

(defun string->function (name &key (readtable-case (readtable-case *readtable*))
                              (package *package*))
  (let ((effective-name (map 'string
                             (ecase readtable-case
                               ((:upcase)
                                #'char-upcase)
                               ((:downcase)
                                #'char-downcase)
                               ((:preserve)
                                #'identity)
                               ((:invert)
                                (lambda (c)
                                  (cond
                                   ((upper-case-p c)
                                    (char-downcase c))
                                   ((lower-case-p c)
                                    (char-upcase c))
                                   (t c)))))
                             name)))
    (multiple-value-bind (s status) (find-symbol effective-name package)
      (unless status
        (error "no symbol for ~S (from ~S)" effective-name name))
      (unless (fboundp s)
        (error "no function for ~S (from ~S, originally ~S)" s effective-name name))
      (symbol-function s))))

现在

> (string->function "+")
#<Function + 80E004F189>

> (eql (string->function "+") #'+)
t

> (string->function "string->function")
#<Function string->function 80200011E9>

> (string->function "unknown")

Error: no symbol for "UNKNOWN" (from "unknown")

> (string->function "*print-case*")

Error: no function for *print-case* (from "*PRINT-CASE*", originally "*print-case*")