How to generate adder function in lisp?

Question

I need to make a function that generates functions like this:

(defun add3 (val) (+ 3 val))

It should be usable like this:

(setq adder7 nil)
(fset 'adder7 (make-adder 7))
(adder7 3)
     => 10

Ideally, function make-adder should return lambdas that do not have any symbols inside other then their parameter, for example:

(make-adder 7)
      => (lambda (val) (+ 7 val))

Update

I have tried following naive realisation:

(defun make-adder (n)
  (lambda (x) (+ n x)))

But that generates a lambdda that contains free symbol (not numeric constant!) and its usage fails.

(defalias 'add1 (make-adder 1))
(add1 2)
       => Debugger error void-variable n

(let ((n 5))
  (add1 2))
       => 7

Which is not what i want to get at all.

Answer 1

Emacs

Emacs relies on dynamic scoping by default. That's why the n symbol inside the returned lambda refers to an unbound variable. Either you toggle lexical scoping or you build a lambda form with the current value of n, as follows:

(defun make-adder (n)
  `(lambda (x) (+ ,n x)))

And then:

(defalias  'add1 (make-adder 1))
(add1 3)
=> 4 

Common Lisp

I originally thought the question was about Common Lisp (fset should have given me a hint), where you only have to do this:

(defun make-adder (n)
  (lambda (x) (+ n x)))

Your function takes a n and returns an anonymous function, which takes another parameter x and produces the result. But wait, make-adder is just a special case of partially applying some arguments to a function (see this question for details on the distinction between currying and partial application). The general approach to partially apply functions is:

(defun partial (function &rest partial-args)
  (lambda (&rest args)
    (apply function (append partial-args args))))

For example:

(let ((3+ (partial #'+ 3)))
  (funcall 3+ 7))
=> 10

Answer 2

coredump-'s answer address the right issue: emacs's dynamic scoping means that you no longer have access to the value of n that was in scope when your adder function was called. Rather than using backquote to construct a function with the constant injected, though, I think that it might be more descriptive to use emacs's lexical-let, since what you're looking for is a lexical closure. Just to be clear, here's the behavior you're getting now, with dynamic scoping:

(defun dyn-adder (n)
  (lambda (x)
    (+ n x)))

(funcall (dyn-adder 3) 5)
;; (void-variable n) error...

And here's how you can use lexical-let to get an actual lexical closure:

(defun lex-adder (n)
  (lexical-let ((n n))
    (lambda (x)
      (+ n x))))

(funcall (adder 3) 5)
;; 8

There isn't much difference between the backquote solution and lexical-let in defining a simple adder, but there are contexts where it's important to actually have a variable that you're referencing. E.g., if you wanted to create an accumulator, you'd need that local state. With dynamic scoping you'd get the same void-variable error:

(defun dyn-accumulator (n)
  (lambda (x)
    (incf n x)))

(let ((acc (dyn-accumulator 5)))
  (funcall acc 3)
  (funcall acc 8))
;; (void-variable n) error...

With a backquote approach, you'll get a different error. (I'm not sure I'm doing this correctly, though. I think that the error I'm getting is from trying to funcall a list, not from the function having a constant in it. But in any case, it should be clear that (lambda (x) (incf 5 x)) won't work, because you can't increment the constant value.)

(defun bq-accumulator (n)
  `(lambda (x)
     (incf ,n x)))

(let ((acc (bq-accumulator 5)))
  (funcall acc 3)
  (funcall acc 8))
;; wrong type argument error...

But with lexical-let, we have a real variable that we can modify:

(defun lex-accumulator (n)
  (lexical-let ((n n))
    (lambda (x)
      (incf n x))))

(let ((acc (lex-accumulator 5)))
  (funcall acc 3)
  (funcall acc 8))
;; 16

I learned about and described lexical-let more in an answer I wrote for Emacs lisp: why does this sexp cause an invalid-function error?; that may be useful information here, as well.