Question

? :

我目前这样做：

for (int index = 0; index < alpha.length; index++) {
    alpha[index] = index * ((index < 25) ? index : 3);
}

对函数名称进行硬编码似乎不是一个好习惯。有关实施{ "_index": "people", "_type": "employee", "_id": "8725", "_source": { "id": 43470, "firstName": "John", "lastName": "Smith", "groups": [ { "id": 345, "name": "Developers" }, { "id": 75432, "name": "Scrummasters" }, { "id": 5789, "name": "UX" } ] } }, { "_index": "people", "_type": "employee", "_id": "8726", "_source": { "id": 43471, "firstName": "Fred", "lastName": "Bloggs", "groups": [ { "id": 474, "name": "Developers" }, { "id": 824, "name": "Admins" } ] } }或(defun foo () (send-to-debug-log "Error. Function terminated." (get-current-function-name)))。

的任何建议

Answer 1

使用包which-function-mode（在melpa中），您可以以编程方式调用函数

(Which-function)

更多信息：

Answer 2

这里似乎运作得很好：

(defun calling-function ()
  (let ((n 6) ;; nestings in this function + 1 to get out of it
        func
        bt)
    (while (and (setq bt (backtrace-frame n))
              (not func))
        (setq n (1+ n)
              func (and bt
                      (nth 0 bt)
                      (nth 1 bt))))
    func))

如果你在lambda中调用它，你就得到了整个lambda。它适用于apply，funcall和eval。

我认为最简单，最健壮的方法就是明确地编写函数名称，正如您现在所做的那样。

Answer 3

您可以尝试以下方式：

(defconst my-defun-macros '(defun cl-defun defmacro cl-defmacro))

(defun my-add-defun-name (orig-fun name &rest args)
  (let* ((my-current-defun-name (if (consp name) (cadr name) name))
         (macroexpand-all-environment
          (cons (cons 'my-get-current-function-name
                      (lambda () my-current-defun-name))
                macroexpand-all-environment))
         (code (apply orig-fun name args)))
    (macroexpand-all code macroexpand-all-environment)))

(defmacro my-get-current-function-name ()
  "Return current defun's name.
Only works within the advised macros in `my-defun-macros'."
  (error "my-get-current-function-name used outside of a defun"))

(dolist (m my-defun-macros)
  (advice-add m :around #'my-add-defun-name))

有没有办法获取当前函数的函数名？

4 个答案: