Re^4: 5.42: Does m// toss a string around?

(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
[download]

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Re^5: 5.42: Does m// toss a string around? by ikegami (Patriarch) on Jan 30, 2026 at 18:59 UTC
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` [download] 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.	[reply] [d/l] [select]
Re^6: 5.42: Does m// toss a string around? by Anonymous Monk on Jan 30, 2026 at 22:37 UTC
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; $_ .= "a"; 1 while /./g' real 0m35.868s user 0m35.857s sys 0m0.008s` [download]	[reply] [d/l]
Re^6: 5.42: Does m// toss a string around? by Anonymous Monk on Jan 30, 2026 at 22:47 UTC
And so as not to mention 5.18 again, and I'm not advocating the use of `s///` as below but FWIW: `$ 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` [download]	[reply] [d/l] [select]


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