Sorry for the late response - missed this for a while.
That's still much more slow than not doing it - slightly modifying your
example,:
(progn
(setq start (current-time))
(let ((row 0) (log (time-add (current-time) '(0 1 0 0))))
(while (< row 6543210)
(setq row (1+ row))
(when (time-less-p log (current-time))
(setq log (time-add (current-time) '(0 1 0 0)))
(message "row %d" row))))
(setq end (current-time))
(print (time-subtract end start)))
prints (0 43 386499 0) on my computer.
Removing the when clause:
(progn
(setq start (current-time))
(let ((row 0) (log (time-add (current-time) '(0 1 0 0))))
(while (< row 6543210)
(setq row (1+ row))))
(setq end (current-time))
(print (time-subtract end start)))
Results in:
(0 1 277641 0)
So adding the logging here slows it down by about 43x - It doesn't seem
worth it.
On Sun, Aug 17, 2014 at 6:39 AM, Michael Brand <[email protected]>
wrote:
> Hi Nathaniel
>
> On Thu, Aug 7, 2014 at 4:57 PM, Nathaniel Flath <[email protected]>
> wrote:
> > I'd be fine with displaying every
> > second, but I don't see a good way of doing this - do you have any
> > suggestions?
>
> I thought of something like in this example:
>
> (let ((row 0) (log (time-add (current-time) '(0 1 0 0))))
> (while (< row 6543210)
> (setq row (1+ row))
> (when (time-less-p log (current-time))
> (setq log (time-add (current-time) '(0 1 0 0)))
> (message "row %d" row))))
>
> Michael
>
