how to defun a function within a defun?

Question

for example, I pass the function name to another funciton

(personal-function 'func-name '(attr1 attr2 ...))

and what I want to do is

(defun personal-function (func-name)
     (defun func-name '(attr1 attr2 ...) (dosomething)))

however, it said I can't defun with a symbol.. what should I do?

Thanks so much!

Answer 1

You could use a lambda

http://www.n-a-n-o.com/lisp/cmucl-tutorials/LISP-tutorial-21.html

If you're trying to make a new globally-accessible function inside a function, I don't think the language's grammar allows for that. If you create a lambda, you can initialize a variable to this lambda value and pass the variable around to your functions. In Common LISP, you can call (functionp x) to determine if a variable is a function before you try to call it.

Answer 2

Solve it as follows:

e.g1

(defun create-function(a1)
  (defun plus-function(x) (+ x a1)))

(create-function 2) -> PLUS-FUNCTION
(plus-function 3) ->5

e.g2

(setf (symbol-function 'printx) #'(lambda (x) (print x)))

(printx '(1 2 3)) -> (1 2 3)

---

Previously I also had the same problem when I defined the function.

Example:

(defun test-function(fn)
            (defun fn ((lambda() (print "aaa")))))

After I run

(test-function 'aaafunction)

The evaluation result is

FN

It does not return a function named "aaafunction"...

To the person who downvote my answer: We are newbies of Lisp, but we are working hard to learn knowledge, and you are not so respectful.

Answer 3

I use defmacros for metaprogramming, for example, this emacs lisp example.

Answer 4

Use

(setf (symbol-function my-symbol) some-function)

create a new function with

(compile nil (list 'lambda args body))

where args and body have meaningful values.