How to defun a function within a defun?

Question

For example, I pass the function name to another function

(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?

Cycad

defun does not evaluate the name, the arglist or the body. all have to be literal values. So your code has two problems. 1) it will define a function func-name. 2) a quoted arglist will also not do what you expect. – Wimble

Answer 1

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.

Wimble

Can't understand: what to use, setf or compile, or some combination of both (what does it looks like then?). You didn't actually provide an example of function definition inside other function definition in the spirit of example from original post. – Nottage

@Artem Pelenitsyn: I would first create the function object and then set the symbol function to it. – Wimble

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.

Pilot

defun returns the function name, not the function: cs.cmu.edu/Groups/AI/html/hyperspec/HyperSpec/Body/… – Scorekeeper

defun does not evaluate the name – Wimble

So how do I force it to evaluate this "fn"? – Pilot

I guess the downvote is because the answer is not really useful---it does not answer the question. Your effort to learn is appreciated, but that is not what gets you votes here. This is not a forum. – Zobias

So now I have given the solution, the downvote should be canceled right? But you know what, those people who downvote will not visit this question again. – Pilot

Answer 3

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.

Milkmaid

thanks a lot. however, my interpreter tells me define is not a function...? "Error: attempt to call `DEFINE' which is an undefined function." – Cycad

are you using Common LISP? Scheme? – Milkmaid

see this for the use of functionp lispworks.com/documentation/HyperSpec/Body/f_fnp.htm... the combination of lambda and functionp should do what you want. – Milkmaid

i don't understand where you go the notion to try to call a function called define. Maybe it was in the first link i posted accidentally. Have you looked at the new link in the edited question? – Milkmaid

oh, I saw the edited one! but could you plz tell me how to init a var to a lambda value? thanks so much!!!!! – Cycad

Answer 4

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

Phocaea

Answer 5

Using a macro will be a good choice. such as (defmacro personal-function (func-name) `(defun ,func-name '(attr1 attr2 ...) (dosomething))) please have a try, hope that helps you.

Drambuie