Re: Ways to delete start of string

The speed depends on the length of the string.

Here's my modified benchmark:

#!/perl/bin/perl
#
# bench.pl --
use strict;
use warnings;
use diagnostics;
use Benchmark qw(:all);

my $str = '|0' x (shift @ARGV || 100_000);

sub caseone {
    $_ = $str;
    s/.//;
}

sub casetwo {
    $_ = $str;
    substr($_,0,1) = '';
}

sub casethree {
    $_ = $str;
    $_ = reverse $_;
    chop;
    $_ = reverse $_;
}

sub casefour {
    $_ = $str;
    $_ = substr($_,1);
}

cmpthese(-2, {
    'regex' => 'caseone',
    'substr_mod' => 'casetwo',
    'reverse' => 'casethree',
    'susbtr_copy' => 'casefour',
});
[download]

# string length 2 * 10:
                 Rate     reverse       regex  substr_mod susbtr_copy
reverse      797484/s          --        -29%        -30%        -41%
regex       1125463/s         41%          --         -1%        -16%
substr_mod  1131927/s         42%          1%          --        -16%
susbtr_copy 1344489/s         69%         19%         19%          --


# string length 2 * 100_000:
              Rate     reverse susbtr_copy       regex  substr_mod
reverse     1385/s          --        -51%        -60%        -80%
susbtr_copy 2847/s        106%          --        -17%        -58%
regex       3437/s        148%         21%          --        -49%
substr_mod  6771/s        389%        138%         97%          --

# with 2 * 1e7:
              Rate     reverse susbtr_copy       regex  substr_mod
reverse     8.61/s          --        -54%        -58%        -81%
susbtr_copy 18.7/s        117%          --         -8%        -58%
regex       20.4/s        137%          9%          --        -54%
substr_mod  44.6/s        418%        139%        119%          --
[download]

The speed doesn't really depend on the perl version (I tried 5.8.8 and 5.10.0).

Comment on Re: Ways to delete start of string Select or Download Code

Replies are listed 'Best First'.
Re^2: Ways to delete start of string by hsmyers (Canon) on May 24, 2008 at 20:25 UTC
'reverse' seems fairly intuitive, longer the string the greater the work (times two in fact). 'regex' looks like it is seeking constant time of sorts. I'm clueless on the flip flop with 'substr_mod' and 'substr_copy'. --hsm "Never try to teach a pig to sing...it wastes your time and it annoys the pig."	[reply]
Re^3: Ways to delete start of string by moritz (Cardinal) on May 24, 2008 at 20:39 UTC
If you really want to know, look in `pp.c` in the perl source, in blead it's in lines 3055 to 3223. I skimmed it quickly and still have no clue how it does things - too many macros, too little knowledge from my part.	[reply] [d/l]
Re^4: Ways to delete start of string by hsmyers (Canon) on May 24, 2008 at 22:33 UTC
Hmmm... I'm perfectly willing to let sleeping curiosities lie. --hsm "Never try to teach a pig to sing...it wastes your time and it annoys the pig."	[reply]
Re^2: Ways to delete start of string (OOK) by ikegami (Patriarch) on May 27, 2008 at 03:14 UTC
The time needed to perform `$x =~ s/.//;` and `substr($x,0,1) = '';` is not related to the length of $x thanks to the "OOK" optimization. Instead of allocating a new buffer, copying the the string to the new buffer minus the leading char, assigning the new buffer to the variable and freeing the old buffer those two operations increment the pointer to the buffer in the variable, assign 1 to the IV slot* of the variable if the OOK flag is off or increment the IV slot* of the variable if the OOK flag is on, and turn on the OOK flag The string is never copied. You can see this in effect in the following snippet: `>perl -MDevel::Peek -e"my $x='abcdef'; Dump($x); substr($x,0,1)=''; Du +mp($x);" SV = PV(0x226104) at 0x2252e8 REFCNT = 1 FLAGS = (PADBUSY,PADMY,POK,pPOK) PV = 0x182ca64 "abcdef"\0 CUR = 6 LEN = 8 SV = PVIV(0x227134) at 0x2252e8 REFCNT = 2 FLAGS = (PADBUSY,PADMY,POK,OOK,pPOK) IV = 1 (OFFSET) PV = 0x182ca65 ( "a" . ) "bcdef"\0 CUR = 5 LEN = 7` [download] If POK is true and OOK is false, then string start = PV string length = CUR number bytes allocated = LEN start of buffer = PV If POK is true and OOK is true, then string start = PV string length = CUR number bytes allocated = LEN + IV start of buffer = PV - IV * — Nicholas Clark recently made a change to Perl so that the chopped bytes are used instead of the IV slot. That hasn't appeared in any Perl release yet. Update: Changed `substr($x,0,1,'');` to `substr($x,0,1)='';` since the rest of the thread used the latter.	[reply] [d/l] [select]


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