Perl: speed of s/// with variable in replacement-CodePudding

This:

while($read=sysread(STDIN,$buf,32760)) {
    $buf=~s/\r/posttag\rpretag\t/go;
    $buf=~s/\n/posttag\npretag\t/go;
    syswrite(STDOUT,$buf);
}

delivers ~200 MB/s on my system.

This:

my $pretag = "pretag";
my $posttag = "posttag";
while($read=sysread(STDIN,$buf,32760)) {
    $buf=~s/\r/$posttag\r$pretag\t/go;
    $buf=~s/\n/$posttag\n$pretag\t/go;
    syswrite(STDOUT,$buf);
}

delivers ~100 MB/s on my system.

Why?

I thought that when I had used /o it should not matter if the content is a variable or a fixed string. Is there an easy way I can get the speed of the first?

CodePudding user response：

Are you sure that regexp recompilation is to blame? You can get information about what perl is doing with your regexps using

use re 'debug';

In second case you introduce interpolated string in s//HERE/ and perl interpolates it with each iteration.

Try to rewrite like

my $pretag = 'pretag';
my $posttag = 'posttag';
my $first_replace = "$posttag\r$pretag\t";
my $second_replace = "$posttag\n$pretag\t";
while($read=sysread(STDIN,$buf,32760)) {
    $buf=~s/\r/$first_replace/go;
    $buf=~s/\n/$second_replace/go;
    syswrite(STDOUT,$buf);
}

CodePudding user response：

Per the perlop documentation for the /o flag, emphasis added:

"s/PATTERN/REPLACEMENT/msixpodualngcer"

If you want the pattern compiled only once the first time the variable is interpolated, use the "/o" option.

Therefore /o is only relevant to the pattern, not the replacement. Even then some programmers consider /o to be a problem, and will not use it, with the typical sentiment running along the lines of:

< pragma-> perlre#Modifiers: o - pretend to optimize your code, but actually introduce bugs

If I had to interpolate the Right-Hand-Side or replacement in advance, I'd probably use Sub::Quote or similar to turn the RHS into static strings.