Re: Making sure I'm not checking email (or doing other things) too often

[Marcin Borkowski <mbork@mbork.pl>]

On 2016-02-15, at 15:40, Michael Heerdegen <michael_heerdegen@web.de> wrote:

> Factoring out the greatest common divider would give you something like
> this:
>
> --8<---------------cut here---------------start------------->8---
> -*- lexical-binding: t -*-
>
> (defun stopwatch ()
>   "Return a fresh stopwatch.
> This is a function accepting zero arguments that upon each call
> will return the time difference from its last call in seconds.
> When called the first time it will return nil."
>   (let ((last-invocation nil))
>     (lambda ()
>       (prog1 (and last-invocation
>                    (time-to-seconds (time-subtract (current-time) last-invocation)))
>          (setq last-invocation (current-time))))))
> --8<---------------cut-here---------------end--------------->8---
>
> With that, one could make your advice functions reference and use such a
> stopwatch object (each an individual one, of course).

Thanks.  This is my version:

(defun create-stopwatch ()
  "Return a closure which returns time (in seconds) since its
last invocation every time it is called."
  (let (last-invocation-time (current-time))
    (lambda ()
      (prog1
	  (time-to-seconds (time-subtract (current-time) last-invocation-time))
	(setq last-invocation-time (current-time))))))

I don't claim it to be better, but I noticed a strange phenomenon.  When
I invoke the "stopwatch object" (closure, in fact) for the first time,
I get negative result.  Why is that so?

> Regards,
>
> Michael.

Best,

-- 
Marcin Borkowski
http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski
Faculty of Mathematics and Computer Science
Adam Mickiewicz University