Lisp quote work internally

Question

How does lisp quote work internally? For example:

(quote (+ 1 (* 1 2)) )

seems to be equivalent to

(list \'+ 1 (li         


        

Answer 1

quote does nothing more than return its argument unevaluated. But what is an unevaluated argument?

When a Lisp program is defined, it is either read from textual source into s-expression form or constructed directly in terms of s-expressions. A macro would be an example of generating s-expressions. Either way there is a data structure comprising (mostly) symbols and conses that represents the program.

Most Lisp expressions will call upon evaluation and compilation machinery to interpret this data structure as terms in a program. quote is treated specially and passed these uninterpreted symbols and conses as its argument. In short, quote does almost nothing - the value it returns already exists and is simply passed through.

You can observe the difference between passing through and fresh construction by using eq to test the identity of the return value of quote:

(defun f () '(1 2))

(defun g () (list 1 2))

(eq (f) (f)) => T
(eq (g) (g)) => NIL

As you can see, quote returns the same conses each time through.