emacs - 如何将函数作为参数传递给 elisp？

Question

我试图在 elisp 中将一种方法传递给另一种方法，然后让该方法执行它。这是一个例子：

(defun t1 ()
  "t1")

(defun t2 ()
  "t1")

(defun call-t (t)
  ; how do I execute "t"?
  (t))

; How do I pass in method reference?
(call-t 't1)

Answer 1

首先，我不确定命名你的函数t是否有帮助，因为 't' 被用作 lisp 中的真值。

也就是说，以下代码对我有用：

(defun test-func-1 ()  "test-func-1"
   (interactive "*")
   (insert-string "testing callers"))

(defun func-caller (callee)
  "Execute callee"
  (funcall callee))

(func-caller 'test-func-1)

请注意“funcall”的使用，它会触发实际的函数调用。

Answer 2

Emacs Lisp 手册中“ §13.7 Anonymous Functions ”末尾的注释说，您可以引用函数而不是向字节编译器发出符号始终命名函数的信号。#''

Answer 3

Above answers are okey, but you can do something more interesting with defmacro, wich evaluates functions later for some reason:

(defun n1 ()
  "n1")

(defmacro call-n (n)
  (apply n))

(call-n (n1))

A practical example with a for loop that takes any amount of functions and their arguments:

(defmacro for (i &optional i++ &rest body)
  "c-like for-loop"
  (unless (numberp i++) (push i++ body) (setq i++ 1))

  (while (/= i 0)
    (let ((args 0))
      (while (nth args body)
    (apply (car (nth args body))
           (cdr (nth args body)))
    (setq args (1+ args))))
    (setq i (- i i++))
    )
  )