CODEWITHSUNDEEP
lispcommon-lisp
sterling
sterling

Reputation: 109

How is my LISP function an unbound variable?

I define a function in LISP, and it defines correctly. But whenever I try to call it, I get an error saying "The variable FACTORIAL is unbound."

I have tried this on both OS X and Windows 7, on LispWorks and Allegro. The function is - 

(defun factorial (x)
   (if (= 1 x) 1
       (* x factorial (- 1 x))))

Any help is appreciated.

Upvotes: 10

Views: 20911

Answers (4)

lkahtz
lkahtz

Reputation: 4796

You need to bind all variables and function calls you intend to use with brackets unless you want them to be symbols. About unbound errors, Paul Graham has a good example in his book: Ansi Common Lisp

One of the most common complaints you'll hear from Lisp is that a symbol has no value or is unbound. Several distinct problems show themselves in this way. Local variables, like those established by let and defun, are valid only within the body of the expression where they are created. So if we try to refer to such a variable outside the let that creates it:

> (progn (let ((x 10))
                    (format t "Here x = ~A.~%" x))
                  (format t "But now it's gone...~%") x)

Upvotes: 0

mathk
mathk

Reputation: 8143

To complete the answer of @Isaac Hodes, this show you that there is clearly 2 namspace for function and variable in CL. You wouldn't have the same error if you were in scheme. You can read more here.

Upvotes: 1

Isaac
Isaac

Reputation: 16392

In the third line of your code, you're multiplying x times factorial times 1-x.

The first thing to notice is factorial isn't a variable: it's a function. As Common-Lisp is a Lisp-2, factorial isn't bound as a variable at all–it's bound as a function.

You need to be calling the factorial function on one less than x, not x less than one.

So: 

(defun factorial (x)
   (if (= 1 x) 1
       (* x (factorial (- x 1)))))

…should do it.

Upvotes: 17

Greg Hewgill
Greg Hewgill

Reputation: 994589

It looks like you're missing a set of parentheses:

(defun factorial (x) 
   (if (= 1 x) 1 
       (* x (factorial (- 1 x)))))

Without the () around factorial, Lisp thinks you're referring to a variable instead of a function.

Upvotes: 5

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