refaddr返回的值是永久的吗？

Question

根据Scalar::Util's documentation，refaddr的工作原理如下：

my $addr = refaddr( $ref );

     

如果引用$ ref，则引用值的内部存储器地址将作为普通整数返回。否则返回undef。

但是，这并不能告诉我$addr是永久性的。引用的refaddr会随着时间的推移而改变吗？例如，在C中，运行realloc可能会更改存储在动态内存中的内容的位置。这与Perl 5类似吗？

我问，因为我想制作一个inside-out object，我想知道refaddr($object)是否能成为一把好钥匙。例如，在XS中编程时，它似乎最简单。

Answer 1

首先，不要重新发明轮子;使用Class::InsideOut。

---

这是永久性的。必须是，否则以下会失败：

my $x;
my $r = \$x;
... Do something with $x ...
say $$r;

Scalars有一个＆＃34; head＆＃34;在固定的位置。如果SV需要升级（例如保持字符串），它就是第二个存储块，称为＆＃34; body＆＃34;那会改变。字符串缓冲区是第三个内存块。

$ perl -MDevel::Peek -MScalar::Util=refaddr -E'
   my $x=4;
   my $r=\$x;
   say sprintf "refaddr=0x%x", refaddr($r);
   Dump($$r);
   say "";

   say "Upgrade SV:";
   $x='abc';
   say sprintf "refaddr=0x%x", refaddr($r);
   Dump($$r);
   say "";

   say "Increase PV size:";
   $x="x"x20;
   say sprintf "refaddr=0x%x", refaddr($r);
   Dump($$r);
'

refaddr=0x2e1db58
SV = IV(0x2e1db48) at 0x2e1db58             <-- SVt_IV variables can't hold strings.
  REFCNT = 2
  FLAGS = (PADMY,IOK,pIOK)
  IV = 4

Upgrade SV:
refaddr=0x2e1db58
SV = PVIV(0x2e18b40) at 0x2e1db58           <-- Scalar upgrade to SVt_PVIV.
  REFCNT = 2                                    New body at new address,
  FLAGS = (PADMY,POK,IsCOW,pPOK)                but head still at same address.
  IV = 4
  PV = 0x2e86f20 "abc"\0                    <-- The scalar now has a string buffer.
  CUR = 3
  LEN = 10
  COW_REFCNT = 1

Increase PV size:
refaddr=0x2e1db58
SV = PVIV(0x2e18b40) at 0x2e1db58
  REFCNT = 2
  FLAGS = (PADMY,POK,pPOK)
  IV = 4
  PV = 0x2e5d7b0 "xxxxxxxxxxxxxxxxxxxx"\0   <-- Changing the address of the string buffer
  REFCNT = 2                                    doesn't change anything else.
  CUR = 20
  LEN = 22