How are function parameters stored in lisp?

Question

I assumed that values passed into a lisp function are assigned to a quote matching the name of the parameter. However, I was surprised that this:

(defun test (x) (print (eval 'x)))
(test 5)

doesn't work (the variable x is unbound). So if parameters aren't stored as symbols in the function, what exactly IS x in this example? Is there a way to access parameters from a symbol matching the parameter name?

More context: What I would like to do is something like this:

defun slice (r1 c1 r2 c2 board)
  (dolist (param '(r1 c1 r2 c2))  ;adjust for negative indices
    (if (< (eval param) 0)
      (set param (+ (length board) (eval param)))))
        ;Body of function

Basically, I want to iterate through the first four parameters and make an adjustment to any of their values if they are < 0. Of course, I could do a let and have an individual line for each parameter, but considering I'm doing the same thing for each of the four parameters this seemed cleaner. However, I get the error that the variable R1 is unbound.

Answer 1

That's basically how lexical binding works: the variable name gets replaced within the lexical scope with a direct reference to where the variable's value is stored. Binding the variable name's symbol-value is only done for dynamic variable which you can declare with special.

One way to avoid repeating yourself would be a macro:

(defmacro with-adjusting ((&rest vars) adjust-value &body body)
  `(let ,(loop for var in vars
               collect `(,var (if (minusp ,var)
                                (+ ,var ,adjust-value)
                                ,var)))
     ,@body))

(defun slice (r1 c1 r2 c2 board)
  (with-adjusting (r1 c1 r2 c2) (length board)
    ;; function body