Understanding data mode in Lisp, why are these expressions not the same?

Question

Currently I am reading "Land of Lisp". In one of the recent code samples the author gave:

> (eq 'fooo 'FoOo)
T

to prove that the symbols are case-insensitive. A few pages later data mode is formally introduced.

However I fail to really understand the following. eq is a function, so its name is case-insensitive as well. Therefore I should be able to do this:

> (eq 'Eq 'EQ)
T

Great. That worked as expected. But what if I put this into a list in data mode? Keep in mind, I am just experimenting with something that's new for me.

> (eq '(Eq) '(EQ))
NIL
> (eq '('Eq) '('EQ))
NIL

Uhm. Okay? Why is that? I would have expected that if I put the same symbol into two lists, that the lists would be considered equal.

Now the question: does that mean that not the contents of the lists are compared, but the list "objects" themselves? What am I missing?

Answer 1

Symbols are Case-Sensitive

> (eq (print (intern "foo")) (print (intern "FOO")))
|foo|   ; printed 
FOO     ; printed
==> NIL ; returned

The default reader is Case-Converting

(eq (read-from-string "foo") (read-from-string "FOO"))
==> T

However, you can make the reader case-preserving:

(let ((*readtable* (copy-readtable)))
  (setf (readtable-case *readtable*) :preserve)
  (eq (read-from-string "foo") (read-from-string "FOO")))
==> NIL

Please take a look at

• Reader Algorithm
• Effect of Readtable Case on the Lisp Reader

EQ compares for pointer identity

Common Lisp provides 4 equality predicates, you should use the right one for your needs:

(equal '(Eq) '(EQ))
==> T
(equal '('Eq) '('EQ))
==> T