LISP local/global variable assignment

Question

If we define a function something like

(defun foo(x)
  (setf x somevalue))

Is x defined as a local variable or global? using setf/

Answer 1

Actually, x is local to the function, so setf or setqing it only changes the local variable x.

To answer the second part of your question, there is another way to define a local variable other than let:

(funcall (lambda (var1 var2...)
           ... use var1 var2 ...)-
   val1 val2 ...)

In fact, defun could be implemented as

`(setf (symbol-function ,name) (lambda ,args ,@body))

though all the implementations I've checked do more than that.

Also, the symbol-function place allows you to do this:

(setf (symbol-function '+) (function -))

though that's generally a bad idea.

Your second question

What's happening there is that x is local to the scope that contains the defun. It's not a global variable. What you're doing with that defun is creating a closure that "captures" all the lexically-scoped (not created with defvar/defparameter) variables in the surrounding scopes and holds on to them for the future. If you want to inspect the value of x, add another function

(defun get-x () x)

inside the let. To put it another way, x is local to the let that the function is in. The situation you've provided is similar to:

(let ((x 3))
    (print x)
    (let ((y 7))
        (print (list x y))
        (setf x y))
    (print x))

except the inner let is replaced by a defun (which is a lambda).