On Sun, Feb 3, 2019 at 9:36 PM ToddAndMargo via perl6-users
<perl6-us...@perl.org <mailto:perl6-us...@perl.org>> wrote:
Hi All,
If I have a variable of type Buf which 10000000 bytes in it
and I find the five bytes I want, is it faster, slower,
or no difference in speed to overwrite the same variable
with the five bytes? Or is it faster to put the five bytes
from the first variable into a second variable?
Many thanks,
-T
On 2/5/19 8:42 AM, yary wrote:
There are modules to time two pieces of code and show the difference
https://github.com/perl6-community-modules/perl6-Benchy
https://github.com/tony-o/perl6-bench
You can write up the two versions you're thinking of, feed them to the
benchmark module, and show us what you find!
-y
Hi Yary,
Thank you!
Apparently, overwriting the original buffer only change the
structures pointers, which is almost instantaneous.
And you taught me something new today!
-T
<code VarTest.pl6>
#!/usr/bin/env perl6
use Bench;
my IO::Handle $HaystackHandle = open( "/home/temp/procexp64.exe", :bin,
:ro );
my Buf $Haystack = $HaystackHandle.read( 3000000 );
$HaystackHandle.close;
my Buf $Needle;
my $b = Bench.new;
sub Another() { $Needle = $Haystack.subbuf( 0x14FFAC .. 0x145FAC ); }
sub Same() { $Haystack = $Haystack.subbuf( 0x14FFAC .. 0x145FAC ); }
say "first copies to a new variable; second overwrites the same variable";
$b.timethese( 100000, {
first => sub { Another; },
second => sub { Same; },
});
</code VarTest.pl6>
$ VarTest.pl6
first copies to a new variable; second overwrites the same variable
Benchmark:
Timing 100000 iterations of first, second...
first: 0.021 wallclock secs (0.021 usr 0.000 sys 0.021 cpu) @
4676612.262/s (n=100000)
second: 0.000 wallclock secs (0.000 usr 0.000 sys 0.000 cpu)
