Re: Bug: buffer local variables handled wrong [9.0.5 (release_9.0.5-497-g5bc540 @ /home/hs/.emacs.d/lib/org-mode/lisp/)]

["Stefan-W. Hahn" <stefan.hahn@s-hahn.de>]

Mail von Nicolas Goaziou, Sat, 03 Jun 2017 at 22:46:01 +0200:

Good morning,

> >
> > This error comes from org-clone-local-variables, because there the
> > prediction for local variables is always to be a list.
> 
> Fixed. Thank you for the report and the analysis.

Thanks.

Because for me pcase is magic, I tried to understand your corretion of the
code.

To understand it I expanded the pcase form with (macrostep-expand), e.g. in
org-agenda-mode:

,----
| (defun org-agenda-mode ()
| ...
| 	   (mapc #'make-local-variable org-agenda-local-vars)
| 	   (dolist (elem save)
| 	     (if
| 		 (consp elem)
| 		 (let*
| 		     ((x
| 		       (car elem))
| 		      (x
| 		       (cdr elem)))
| 		   (if
| 		       (consp x)
| 		       (let*
| 			   ((x
| 			     (car x))
| 			    (x
| 			     (cdr x)))
| 			 (if
| 			     (null x)
| 			     (let
| 				 ((val x)
| 				  (var x))
| 			       (when
| 				   (and val
| 					(memq var org-agenda-local-vars))
| 				 (set var val)))
| 			   nil))
| 		     nil))
| 	       nil)))
| 	 (setq-local org-agenda-this-buffer-is-sticky t))
| ...
`----

for me it looks not correct, because var and val get expanded to (cdr
(cdr elem)).

Also the second one

,----
| (defun org-clone-local-variables (from-buffer &optional regexp)
| ...
|   (dolist (pair (buffer-local-variables from-buffer))
|     (if
| 	(consp pair)
| 	(let*
| 	    ((x
| 	      (car pair))
| 	     (x
| 	      (cdr pair)))
| 	  (if
| 	      (consp x)
| 	      (let*
| 		  ((x
| 		    (cdr x)))
| 		(if
| 		    (null x)
| 		    (let
| 			((name x))
| 		      (when
| 			  (and
| 			   (not
| 			    (memq name org-unique-local-variables))
| 			   (or
| 			    (null regexp)
| 			    (string-match-p regexp
| 					    (symbol-name name))))
| 			(set
| 			 (make-local-variable name)
| 			 (cdr pair))))
| 		  nil))
| 	    nil))
|       nil)))
`----

here name get expanded to (cdr (cdr pair)).

For me both cases don't look correct or do I missinterpret something?

With kind regards,
Stefan

-- 
Stefan-W. Hahn                          It is easy to make things.
                                        It is hard to make things simple.