Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

I know Perl maintains a pointer (offset), within an SV, of where actual data in PV buffer starts, -- is this "beheading" the reason for the slowdown as demonstrated?

use strict;
use warnings;
use Benchmark 'cmpthese';

my $str = ' ' x 1e5;

sub test1 {
    my $s = shift;
    my $copy = substr $str, 1;
    1 while $copy =~ /./g;
    _:
}

sub test2 {
    my $s = shift;
    substr $s, 0, 1, '';
    1 while $s =~ /./g;
    _:
}

cmpthese -1, {
    test1 => sub { test1( $str )},
    test2 => sub { test2( $str )},
};

#             (warning: too few iterations for a reliable count)
#         Rate test2 test1
# test2 2.79/s    --  -93%
# test1 41.0/s 1372%    --

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Re: Are beheaded strings known to be slow?
by tonyc (Hermit) on Oct 10, 2025 at 00:24 UTC

    I suspect it's because OOK SVs aren't Copy-On-Write-able.

    I did a quick profile, and with test2() the code spends most of it's time in memcpy() called from reg_set_capture_string().

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      That's what it is. Regex matching makes a copy of the scalar for use by $&, $1, etc. It uses the COW mechanism if possible. It doesn't for OOK scalars.

      After one pass of test1's loop: 

      SV = PVMG(0x5bb22b8c3b80) at 0x5bb22b9beae8
  REFCNT = 1
  FLAGS = (SMG,POK,IsCOW,pPOK)    <-- IsCOW: String buffer is shared
  IV = 0
  NV = 0
  PV = 0x5bb22b9ea230 "..."
  CUR = 99999
  LEN = 100001
  COW_REFCNT = 1
  MAGIC = 0x5bb22b9c48f0
    MG_VIRTUAL = &PL_vtbl_mglob
    MG_TYPE = PERL_MAGIC_regex_global(g)
    MG_FLAGS = 0x40
      BYTES
    MG_LEN = -1

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      After one pass of test2's loop: 

      SV = PVMG(0x5bb22b8c3bb0) at 0x5bb22b9cc1f0
  REFCNT = 1
  FLAGS = (SMG,POK,OOK,pPOK)    <-- !IsCOW: String buffer isn't shared
  IV = 0
  NV = 0
  OFFSET = 1
  PV = 0x5bb22ba028e1 ( "\x01" . ) "..."
  CUR = 99999
  LEN = 100001
  MAGIC = 0x5bb22b9c48f0
    MG_VIRTUAL = &PL_vtbl_mglob
    MG_TYPE = PERL_MAGIC_regex_global(g)
    MG_FLAGS = 0x40
      BYTES
    MG_LEN = -1

      [download]

      That means a copy of the string buffer must have been made. By the end of the loop, a copy has been done 99,999 times (once per successful regex match).

      You can run into the same issue with foreign string buffers (e.g. a memory-mapped file). See why is Perl File::Map so slow compared to File::Slurp?

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      Thanks for clarification. I.e. we better be careful with 4-args (or LHS) variant of substr?

      It's funny, Perl rightfully decides (case "0") it's cheaper to move 499_998 initial bytes _up_ than moving the trailing 500_000 bytes _down_ (and so scalar gets OOK-"contaminated"); and it's the opposite (rightfully) for "2" -- scalar stays OOK-free & usable OK with regex engine. When ("1") they are both 499_999, then... -- ??? move the 1st half UP and actually _complicate_ things! (do additional work of setting the OFFSET?)

      use Devel::Peek;
$Devel::Peek::pv_limit = 1;

for ( 0 .. 2 ) {
    my $s = 'a' x 1e6;
    substr $s, 499_998 + $_, 2, 'b';
    Dump $s;
}

__END__

SV = PV(0xdbb198) at 0x26f8ae0
  REFCNT = 1
  FLAGS = (POK,OOK,pPOK)
  OFFSET = 1
  PV = 0x2954309 ( ""... . ) "a"...\0
  CUR = 999999
  LEN = 1000001
SV = PV(0xdbb198) at 0x26f8ae0
  REFCNT = 1
  FLAGS = (POK,OOK,pPOK)
  OFFSET = 1
  PV = 0x2954309 ( ""... . ) "a"...\0
  CUR = 999999
  LEN = 1000001
SV = PV(0xdbb198) at 0x26f8ae0
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x2954308 "a"...\0
  CUR = 999999
  LEN = 1000002

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      That's unfortunate... I have code that is using regexes on File::Map buffers under the assumption it was an efficient way to scan through a file. Maybe there's room for one more optimization that checks the length of what needs copied vs. the length of the source scalar? In the example, each iteration only needs to put a single character in $&.
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        You're forgetting about $` and $'. Between $`, $' and $&, the entire string is covered.

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Re: Are beheaded strings known to be slow?
by LanX (Saint) on Oct 09, 2025 at 18:05 UTC
    first observation test1 operates directly on $str not $s

    then you should measure how fast those substrs operate without the regex and if they really always deal with strings of the same length.

    Cheers Rolf
    (addicted to the Perl Programming Language :)
    see Wikisyntax for the Monastery

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      first observation test1 operates directly on $str not $s

      ouch, thanks, I blundered when making SSCCE, s/str/s/ doesn't make any difference

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Re: Are beheaded strings known to be slow?
by ikegami (Patriarch) on Oct 09, 2025 at 18:16 UTC

    Weird.

    There shouldn't be any performance penalty to reading an OOK ("beheaded") scalar. Rather than adding an offset each time the string it accessed, the pointer to the string is changed to point beyond the "head" when the string is "beheaded". 

    use Devel::Peek qw( Dump );
my $s = "abcdef";
$s .= "g";  # Trigger COW
Dump( $s );
substr( $s, 0, 1, "" );
Dump( $s );

    [download]
    SV = PV(0x59d752a54ee0) at 0x59d752a830d0
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x59d752a6b2e0 "abcdefg"\0
  CUR = 7
  LEN = 16
SV = PV(0x59d752a54ee0) at 0x59d752a830d0
  REFCNT = 1
  FLAGS = (POK,OOK,pPOK)
  OFFSET = 1
  PV = 0x59d752a6b2e1 ( "\x01" . ) "bcdefg"\0
  CUR = 6
  LEN = 15

    [download]

    Before the beheading, the string started at 0x59d752a6b2e0 and used 7 bytes of 16.

    After the beheading, the string started at 0x59d752a6b2e1 and used 6 bytes of 15.

    In both cases, the same memory block starting at 0x59d752a6b2e0 is used. But because the PV, CUR and LEN fields were adjusted, reading from the scalar is identical whether it's an OOK ("beheaded") scalar or not.

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