Re^5: 5.42: Does m// toss a string around?

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
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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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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]