[SOLVED] In Scheme or STk, a function will be shown as a procedure or closure, but why does LISP give an error?

In Scheme or STk, a function will be shown as a procedure or closure, but why does LISP give an error?

On Ubuntu, if I run MIT-Scheme, it will show a function as a procedure:

1 ]=> (define (sq x) (* x x))

;Value: sq

1 ]=> (sq 3)

;Value: 9

1 ]=> sq

;Value 11: #[compound-procedure 11 sq]

and Berkeley's STk will show sq as a closure:

STk> (define (sq x) (* x x))
sq
STk> (sq 3)
9
STk> sq
#[closure arglist=(x) b73fab48]

How come in Lisp (Common Lisp clisp), when I do the same thing, it will give me an error instead, and how can I show a function as a value (first class value / object)?

[1]> (defun sq(x) (* x x))
SQ
[2]> (sq 3)
9
[3]> sq

*** - SYSTEM::READ-EVAL-PRINT: variable SQ has no value
The following restarts are available:
USE-VALUE      :R1      Input a value to be used instead of SQ.
STORE-VALUE    :R2      Input a new value for SQ.
ABORT          :R3      Abort main loop

Solution

Common Lisp, unlike Scheme, keeps separate namespaces for variables and function names. Try #'sq in CL. Also google around for 'Lisp1 vs Lisp2' for endless verbiage on the subject.