perl - refaddr 返回的值是永久的吗？

Question

根据 Scalar::Util's documentation , refaddr 是这样工作的:

my $addr = refaddr( $ref );

If $ref is reference the internal memory address of the referenced value is returned as a plain integer. Otherwise undef is returned.

但是，这并没有告诉我 $addr 是否是永久的。引用的 refaddr 会随时间变化吗？例如，在 C 语言中，运行 realloc 可以更改动态内存中存储的内容的位置。这与 Perl 5 类似吗？

我问是因为我想制作一个 inside-out object ，我想知道 refaddr($object) 是否会成为一把好 key 。例如，在 XS 中编程似乎最简单。

Answer 1

首先，不要重新发明轮子；使用 Class::InsideOut .

---

它是永久的。它必须是，否则以下将失败:

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

标量在固定位置有一个“头”。如果 SV 需要升级(例如保存一个字符串)，它是第二个内存块，称为“主体”，将发生变化。字符串缓冲区仍然是第三个内存块。

$ 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