Re: using setq to create lists based on other lists...

[Stephen Berman]

On Sun, 02 Dec 2018 06:21:12 -0500 Barry Margolin <barmar@alum.mit.edu> wrote:

> In article <mailman.5010.1543748027.1284.help-gnu-emacs@gnu.org>,
>  Jean-Christophe Helary <brandelune@gmail.com> wrote:
>
>> I spend most of the day investigating why creating a list with setq was not 
>> "working".
>> 
>> For ex:
>> (setq list0 '(1 2))
>> (setq list1 list0)
>> 
>> If you do
>> 
>> (setcar list0 0)
>> 
>> then for some reason (for which I could not find an explanation in the elisp 
>> reference) the car of list1 also changes, and vice-versa.
>> 
>> Which is totally unexpected since when you do:
>> 
>> (setq list0 0)
>> 
>> list1 does not become 0
>> 
>> I don't suppose that's a bug, but really it ought the be very clearly 
>> documented in the reference. Also, I'd like to know why that's happening.
>
> list0 and list1 both contain references to the same cons. When you use 
> setcar, you're changing the contents of one of the cells in that cons. 
> Since both variables refer to it, the change is visible through either 
> of them.
[...]
> But reassigning the variable doesn't affect the others, because now 
> they're not referring to the same object.

To expand of this, since Jean-Christophe didn't find an explanation of
this behavior of setq in the Lisp reference, but it is in fact
documented:

    Special Form: setq [symbol form]...
     This special form is the most common method of changing a
     variable’s value.  Each SYMBOL is given a new value, which is the
     result of evaluating the corresponding FORM.  The current binding
     of the symbol is changed.

In the above case, the symbol `list1' is given the result of evaluating
`list0', which is the list `'(1 2)'.  So now both `list0' and `list1'
refer to this list, which is a Lisp object; you can see this with `eq',
which returns t if its arguments are the same Lisp object:

(eq list1 list0)
=>
t

That's why setcar affects both `list0' and `list1'.  On the other hand
(setq list0 0) changes the current binding of `list0' to the value of
`0', which is a different object from `'(1 2)', which is still the value
of `list1', so now `list0' and `list1' differ.

Steve Berman