Re: Problem with position and find (cl)

[Thierry Volpiatto]

Florian Beck <abstraktion@t-online.de> writes:

> Marc Tfardy <m-t-o___CUT__IT@web.de> writes:
>
>> Marc Tfardy schrieb:
>>> I try with:
>>> (member '(2) '((1) (2) (3) (4)))
>
> `member' tests the components: ›2‹ ist always equal (in the sense of »eq«) to 
> ›2‹
>
> Compare:
>
> (memq '(2) '((1) (2) (3) (4)))
> (eq 2 2)
> (eq '(2) '(2))
>
>>>
>>> and this gives a expectet results ((2) (3) (4)), but:
>>> (find '(2) '((1) (2) (3) (4)))
>>> or
>>> (position '(2) '((1) (2) (3) (4)))
>>>
>>> returns nil. Why?
>
> Because  the first and the second »(2)« have the same components but are
> *different* lists.
>
>> But this works:
>>
>> (position '(2) '((1) (2) (3) (4)) :test (lambda (x y) (eq (car x) (car
>> y))))
>>
>> Maybe position and find (and maybe some others functions)
>> can not compare lists directly?
>
> They can, but again, the elements are not identical. Compare:
>
> (let* ((x '((1) (2) (3) (4)))
>        (y (cadr x))
>        (z '(2)))
>     (message "x: %s; y:%s" (position y x) (position z x)))
>
> y ist the same object as the cadr of x, z is an entirely new object that
> happens to have the same component.
>
>
position default test is 'eq, try 'equal

,----
| ELISP> (let* ((x '((1) (2) (3) (4)))
|        (y (cadr x))
|        (z '(2)))
|     (message "x: %s; y:%s" (position y x) (position z x :test 'equal)))
| "x: 1; y:1"
`----


-- 
A + Thierry
Pub key: http://pgp.mit.edu