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.

In reply to Re^5: 5.42: Does m// toss a string around? by ikegami
in thread 5.42: Does m// toss a string around? by Anonymous Monk

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