refaddr返回的值是永久的吗？

Question

根据Scalar :: Util的文档,refaddr工作方式如下:

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,这并不能告诉我是否$addr是永久性的.可以在refaddr一段时间内参考的变化？例如,在C中,运行realloc可以改变存储在动态存储器中的东西的位置.这与Perl 5类似吗？

我问,因为我想制作一个由内而外的对象,我想知道是否refaddr($object)能成为一把好钥匙.例如,在XS中编程时似乎最简单.

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;

标量在固定位置有"头".如果SV需要升级(例如,保持字符串),那么它就是第二个被称为"正文"的内存块,它将发生变化.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