in reply to Re^2: 5.42: Does m// toss a string around?
in thread 5.42: Does m// toss a string around?

I think the PV method constructs new string rather than returns numeric value

No.

SvPVX creates an object that provide information about a scalar ($_[0] aka $s), then uses that object's PV method to obtain the address of the string buffer of $s.

I now suspect similar reasons, related to COW, can you explain please?

COW was introduced in 5.20. Before 5.20, hacks which malfunctioned and/or a more expensive alternative had to be used.

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Re^4: 5.42: Does m// toss a string around?
by Anonymous Monk on Jan 29, 2026 at 23:30 UTC

    (OK, the PV method doesn't matter) I thought I got it (s/// implies matching; and $& and friends COW-share, in a way, original buffer; hence replacement operator (after 5.18) always re-allocates) but now I think the following example shows "match and manually replace" doesn't force re-allocation, then why s/// does?

    use Devel::Peek qw( Dump );
$Devel::Peek::pv_limit = 
$Devel::Peek::pv_limit = 10;

$_ = "x" x 99;
$_ .= "x";  # Force unsharing.
Dump( $_ );
/(.+)/;
Dump( $_ );
$_ = 'aaa';
Dump( $_ );

print "\n\n";

$_ = "x" x 99;
$_ .= "x";  # Force unsharing.
Dump( $_ );
s/(.+)/aaa/;
Dump( $_ );


__END__

5.042000
SV = PV(0x1e2951e23e0) at 0x1e29521d298
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x1e295228ad0 "xxxxxxxxxx"...\0
  CUR = 100
  LEN = 101
SV = PV(0x1e2951e23e0) at 0x1e29521d298
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x1e295228ad0 "xxxxxxxxxx"...\0
  CUR = 100
  LEN = 101
SV = PV(0x1e2951e23e0) at 0x1e29521d298
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x1e295228ad0 "aaa"\0
  CUR = 3
  LEN = 101


SV = PV(0x1e2951e23e0) at 0x1e29521d298
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x1e295228ad0 "xxxxxxxxxx"...\0
  CUR = 100
  LEN = 101
SV = PV(0x1e2951e23e0) at 0x1e29521d298
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x1e295215d60 "aaa"\0
  CUR = 3
  LEN = 16

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      So there's really two questions here.

      The first is why wasn't COW used for matching half of the code.

      COW is normally used for the copy for $& and friends, but it wasn't used in this case because there's not enough space in the buffer. 

        CUR = 100
  LEN = 101

      [download]

      I think two bytes are needed. One for the NUL, and one for the COW reference count. Replace $_ .= "x"; with chop;, and you get: 

      SV = PV(0x5deca7958ee0) at 0x5deca7994d68
  REFCNT = 1
  FLAGS = (POK,pPOK)
  PV = 0x5deca7998850 "xxxxxxxxxx"...\0
  CUR = 98
  LEN = 101
SV = PV(0x5deca7958ee0) at 0x5deca7994d68
  REFCNT = 1
  FLAGS = (POK,IsCOW,pPOK)
  PV = 0x5deca7998850 "xxxxxxxxxx"...\0
  CUR = 98
  LEN = 101
  COW_REFCNT = 1
SV = PV(0x5deca7958ee0) at 0x5deca7994d68
  REFCNT = 1
  FLAGS = (POK,IsCOW,pPOK)
  PV = 0x5deca797ad20 "aaa"\0
  CUR = 3
  LEN = 16
  COW_REFCNT = 1

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      The second question is why does the scalar get a new buffer for the substitution.

      I'm guessing the buffer of the scalar is stolen rather than copied for $& and friends, since the expectation is that it will change.

      That means the scalar always gets a new buffer.

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        COW is normally used for the copy for $& and friends, but it wasn't used in this case because there's not enough space in the buffer

        Thanks. I think it follows that appending a character practically poisons a string in Perl (try 1e7 if on modern PC):

        $ time perl -e '$_ = "a" x 1e6; $_ .= "a"; 1 while /./g'

real    0m35.868s
user    0m35.857s
sys    0m0.008s

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        And so as not to mention 5.18 again, and I'm not advocating the use of s/// as below but FWIW:

        $ perlbrew exec --with 5.18.4,5.42.0 time -p perl -e '
>   $_ = 1 x 1e6;
>   1 while s/.$//;
> '
perl-5.18.4
==========
real 0.14
user 0.13
sys 0.00

perl-5.42.0
==========
real 18.95
user 18.94
sys 0.00

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