Re: Symbol's function definition is void: remove-if-not

[Katsumi Yamaoka <yamaoka@jpl.org>]

Ted Zlatanov wrote:
[...]
> Can you use `gnus-remove-if' with the predicate inverted?

It can be used, if the type of a sequence is `list'.  For a non-
list sequence, we can convert it to a list as the function `coerce'
does (see cl-extra.el).  Though the type of a modified sequence has
to be changed again into that of the original (if needed).  However,
it may be hard to make it work for a hash table.  So is the genuine
`remove-if-not'!

> It looks like `remove-if-not' is used in several places actually:

> gnus-art.el:         (remove-if-not pred (mailcap-mime-types))
> gnus-group.el:                          (remove-if-not 'symbolp collection)))
> gnus-score.el:                                            (remove-if-not
> gnus-sum.el:             (remove-if-not 'gnus-valid-move-group-p gnus-active-hashtb)
> gnus-sum.el:             prom (remove-if-not 'gnus-valid-move-group-p gnus-active-hashtb)

The use of `remove-if-not' in gnus-sum.el is a wrong approach since
it doesn't necessarily recognize all the group names in
`gnus-active-hashtb'.  That looks like a normal vector of which
the length is 4096, however it can hold more than 4096 group names
and `remove-if-not' cannot access all of them.  For instance:

;; Make a hash table of which the length is 1.
(let ((hashtable (make-vector 1 0)))
  (length hashtable))
 => 1

;; We can put more than 1 symbol to it,
;; but it still looks like a vector of which the size is 1.
(let ((hashtable (make-vector 1 0)))
  (intern "A" hashtable)
  (intern "B" hashtable)
  (intern "C" hashtable)
  (intern "D" hashtable)
  (list (length hashtable) hashtable))
 => (1 [D])

;; It actually holds four symbols.
(let ((hashtable (make-vector 1 0)))
  (intern "A" hashtable)
  (intern "B" hashtable)
  (intern "C" hashtable)
  (intern "D" hashtable)
  (let (rest)
    (mapatoms (lambda (symbol) (push symbol rest)) hashtable)
    rest))
 => (A B C D)

;; `remove-if-not' recognize the hash table in question as [D],
;; therefore:
(let ((hashtable (make-vector 1 0)))
  (intern "A" hashtable)
  (intern "B" hashtable)
  (intern "C" hashtable)
  (intern "D" hashtable)
  (remove-if-not (lambda (symbol) (string-lessp symbol 'C)) hashtable))
 => []

;; But the expected result of it is like the following, isn't it?
(let ((sequence [A B C D]))
  (remove-if-not (lambda (symbol) (string-lessp symbol 'C)) sequence))
 => [A B]