> (defun benchmark-often ()
> (cl-loop for n from 1 upto 20 do
> (message (format "iteration %d: %f" n (nth 0 (benchmark-run
> (line-number-at-pos (point))))))))
^^^^^^^
Where is this "point" in your tests (I expect the timing to vary
significantly depending on this).
> 1st run:
> iteration 1: 0.001213
> iteration 2: 0.001170
> iteration 3: 0.001170
> iteration 4: 0.001238
> iteration 5: 0.001163
> iteration 6: 0.001153
> iteration 7: 0.000421
> iteration 8: 0.000426
> iteration 9: 0.000322
> iteration 10: 0.000301
> iteration 11: 0.000291
> iteration 12: 0.000292
> iteration 13: 0.000291
> iteration 14: 0.000291
> iteration 15: 0.000295
> iteration 16: 0.000289
> iteration 17: 0.000289
> iteration 18: 0.000288
> iteration 19: 0.000288
> iteration 20: 0.000287
I recommend you don't bother outputting all 20 results: better summarize
it by getting rid of the first test and then giving e.g. the sum or the
median of the rest.
> I'm not allowed to share my employer's source code as a test case, so
> I tried the same procedure with the similarly large DeclBase.h from the
> public LLVM repository. To my surprise, DeclBase.h didn't suffer from any
> performance issues at all.
My crystal ball tells me that DeclBase.h is pure ASCII so byte<->char
conversion is trivial, whereas your file likely contains umlauts and
other disreputable characters.
Here's a similar test case to yours but which builds up its own
artificial buffer with a few non-ascii chars to spice things up:
(with-temp-buffer
(dotimes (i 1000)
(insert "lksajflahalskjdféefawrgfrüegf\n"))
(let ((txtbuf (current-buffer)))
(dotimes (s 4)
(with-temp-buffer
(insert-buffer-substring txtbuf)
(let ((stepsize (lsh 10 (* 4 s))))
(cl-loop for n from (point-min) upto (- (point-max) stepsize)
by stepsize do
(let ((ov (make-overlay n (+ (1- stepsize) n))))
(overlay-put ov 'cquery-sem-highlight t))))
(dotimes (i 4)
(let ((timing
(benchmark-run 1000
(line-number-at-pos
(+ (point-min) (* i (/ (buffer-size) 4)))))))
(message "ols=%S pos=%S/4 time=%.4f (+ %S)"
(/ (buffer-size) (lsh 10 (* 4 s))) i
(car timing) (cdr timing)))
)))))
This gave me (on my top-of-the-line Thinkpad T61 using Debian's `emacs25`):
ols=3000 pos=0/4 time=0.0018 (+ (0 0.0))
ols=3000 pos=1/4 time=6.1074 (+ (0 0.0))
ols=3000 pos=2/4 time=10.6876 (+ (0 0.0))
ols=3000 pos=3/4 time=13.7854 (+ (0 0.0))
ols=187 pos=0/4 time=0.0016 (+ (0 0.0))
ols=187 pos=1/4 time=0.3055 (+ (0 0.0))
ols=187 pos=2/4 time=0.6001 (+ (0 0.0))
ols=187 pos=3/4 time=0.8903 (+ (0 0.0))
ols=11 pos=0/4 time=0.0015 (+ (0 0.0))
ols=11 pos=1/4 time=0.0769 (+ (1 0.006324223))
ols=11 pos=2/4 time=0.1439 (+ (0 0.0))
ols=11 pos=3/4 time=0.2215 (+ (0 0.0))
ols=0 pos=0/4 time=0.0015 (+ (0 0.0))
ols=0 pos=1/4 time=0.0548 (+ (0 0.0))
ols=0 pos=2/4 time=0.1102 (+ (0 0.0))
ols=0 pos=3/4 time=0.1690 (+ (0 0.0))
Stefan
RetroSearch is an open source project built by @garambo | Open a GitHub Issue
Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo
HTML:
3.2
| Encoding:
UTF-8
| Version:
0.7.4