On Sun, Feb 3, 2019 at 9:36 PM ToddAndMargo via perl6-users <perl6-users@perl.org <mailto:perl6-users@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)

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