GNU bug report logs - #4033
23.1; list-colors-display is misleading

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Message #70 received at 4033 <at> emacsbugs.donarmstrong.com (full text, mbox):

From: "Drew Adams" <drew.adams <at> oracle.com>
To: "'Juri Linkov'" <juri <at> jurta.org>
Cc: <4033 <at> debbugs.gnu.org>
Subject: RE: bug#4033: 23.1; list-colors-display is misleading
Date: Tue, 4 Aug 2009 19:25:26 -0700

> > I have had at least one user state that he thought that, since the
> > form shown is #RRGGBB, his colors had only that granularity.  IOW,
> > even if one's system allows colors of the form #RRRRGGGGBBBB (more
> > colors), the color names are translated to hex strings of the form
> > #RRGGBB (fewer colors).
> 
> Could you show a formula that calculates the color granularity?
> 
> We could use it to print colors in the short format #RRGGBB
> for the smaller color space and #RRRRGGGGBBBB otherwise.

What about something like this:

(defun rgb-color-format-for-display ()
  (let ((ncolors  (display-color-cells (selected-frame)))
        (exp      0))
    (while (> (lsh ncolors (- exp)) 1) (setq exp  (1+ exp)))
    (setq exp  (/ exp 12))
    (format "#%%0%dx%%0%dx%%0%dx" exp exp exp)))

For 16777216 colors, that gives #%02x%02x%02x, which seems right.

In `list-colors-print', we would then do this:

(insert (apply 'format (rgb-color-format-for-display)
               (mapcar (lambda (c) (lsh c -8))
                         (color-values (car color)))))

IOW, replace the hard-coded "#%02x%02x%02x" with (rgb-color-format-for-display).

But I guess Jason and Eli are saying that that wouldn't work or wouldn't be
appropriate. It's still not clear to me.

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