I\'m a total Lisp n00b, so please be gentle.
I\'m having trouble wrapping my head around CL\'s idea of an [un-]declared free variable. I would think that:
The Common Lisp HyperSpec (basically its the Common Lisp standard in HTML form) says:
http://www.lispworks.com/documentation/HyperSpec/Body/s_setq.htm
Assigns values to variables.
So SETQ
only assigns values to variables. It does not declare them.
Variable definitions are done globally with DEFVAR
, DEFPARAMETER
, ...
(defparameter *this-is-a-global-dynamic-variable* 'yep)
Variable definitions are done locally with DEFUN
, LET
, LET*
, LOOP
, and many others.
(defun foo (v1 v2)
...)
(let ((v1 10)
(v2 20))
...)
(loop for v1 in '(10 30 10 20)
do ...)
This is basic Lisp and it would be useful to read an introduction. I would recommend:
http://www.cs.cmu.edu/~dst/LispBook/
Above book is free for download.
Additionally the above mentioned Common Lisp Hyperspec provides you with the definitions for Common Lisp and describes the various facilities (DEFUN, LOOP, DEFPARAMETER, ...) in detail.
In Lisp variable declaration may be performed in many ways. The most notable are:
defparameter
and defvar
let
, let*
, multiple-value-bind
, destructuring-bind
, and other binding formsYou can read about their scope in many places as well, for instance in CLtL2.
setq
/setf
are not variable declaration operators, but variable modification operators, as implied by their names.
PS. In interactive mode some implementation would use the DWIM approach and declare the variable as special behind the scenes, if you try to set an undeclared variable, but this is purely for convenience.
In lisp variables may be declared usingdefparameter
ordefvar
.
(defparameter var1 5)
(defvar var2 42)
This results in global (dynamic) variables.
The difference between defvar
and defparameter
is that defvar
does not reinitialize an already existing variable.
Local (lexical) variables are introduced e.g using let
or let*
(which initializes the variables sequentially).
Undeclared free variable means that you have used (here setq
-ed) a variable that it is not bound in the context it is used. It may then be declared for you, but then probably as a global (dynamic) variable. The consequence of this is that if you use undeclared variables with the same name in several functions, you will be referencing the same variable in all of the functions.
Your code can be written like this:
(loop for x from 0 to (- (length str) str-len) do
(let* ((last (+ x str-len)) ; get the last char of substring
(subs (subseq str x last)) ; get the substring
(prod (prod-string subs))) ; get the product of that substring
(if (> prod max) ; if it's bigger than current max, save it
(setq max prod)
(setq max-str subs))))
Using the variable-binding properties of loop, it may also be written as
(loop for x from 0 to (- (length str) str-len)
for last = (+ x str-len)
for subs = (subseq str x last)
for prod = (prod-string subs)
when (> prod max) do
(setq max prod)
(setq max-str subs))