Re^2: Understanding typeglobs and local()

I suppose I knew that, although that didn't occur to me as an explanation.

I guess I'm trying to understand "how it works" on the inside so that I know what behaviour to expect. From what I understand, local ($_); is roughly equivalent to:

my $saved = $_;
$_ = undef;
...
$_ = $saved;
[download]

so foreach saved a pointer to the original $_, which contains 'Sample' as a result of s///. This I understand; it's why $alias1 is unaffected. Yet $alias0 is affected, and I'm not quite sure that I understand why. (This is no different than the local() case.)

Inside the foreach, we alias to $x. This doesn't affect the preexisting aliases, but on the next iteration through the loop our alias still exists. The behaviour changes if the line *_ = *x; is changed to *_ = \$x;, and this is what I was trying to understand.

Hmm, maybe I'm starting to get it now. *x is a 'level higher' than \$x in the internal structure of the symbol table. So foreach is only localizing the scalar (and restoring it at exit - this changes what $_ is after the loop if that change is made), and changes to the typeglob take foreach() completely unawares.

Update: It turns out what was confusing me was that foreach() apparently makes its own copy of the glob for *_, as the alias *_ = *x doesn't change where foreach() puts its alias. *_ = \$x leaves the glob alone, and just modifies the scalar portion; this does get overwritten by foreach()'s aliasing on each iteration (and at loop exit) provided the glob has not been changed.

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Re^3: Understanding typeglobs and local() by ikegami (Patriarch) on Jan 13, 2006 at 03:13 UTC
"Glob" is another name for "symbol table entry". Symbol table entries have multiple fields, one for each data type (scalar, hash, array, etc). `$abc = 'xyz'; print(${{abc}{SCALAR}}, "\n"); # xyz` [download] When you modify a package variable, you modify the data in the appropriate field of the appropriate symbol table entry. `$glob = def = abc; $abc = 'xyz'; print($abc , "\n"); # xyz print($def , "\n"); # xyz print(${$glob }, "\n"); # xyz print(${{abc }{SCALAR}}, "\n"); # xyz print(${{def }{SCALAR}}, "\n"); # xyz print(${{$glob}{SCALAR}}, "\n"); # xyz` [download] Same goes when you localize a package variable. You're localizing the SCALAR field of the symbol table entry. It doesn't matter how you get to the symbol table entry (be it `$_` or `$alias0`, for example), because the value is not associated with the symbol (`$_`), but with the symbol table entry (`${_{SCALAR}}`). `$glob = def = abc; $abc = 'xyz'; print($abc , "\n"); # xyz print($def , "\n"); # xyz print(${$glob }, "\n"); # xyz print(${{abc }{SCALAR}}, "\n"); # xyz print(${{def }{SCALAR}}, "\n"); # xyz print(${{$glob}{SCALAR}}, "\n"); # xyz local $def = 'uvw'; print($abc , "\n"); # uvw print($def , "\n"); # uvw print(${$glob }, "\n"); # uvw print(${{abc }{SCALAR}}, "\n"); # uvw print(${{def }{SCALAR}}, "\n"); # uvw print(${{$glob}{SCALAR}}, "\n"); # uvw` [download] In short, `abc = def;` makes the symbols `abc` and `def` the same. Only changes to the globs can undo this. `local $abc` doesn't affect the glob `abc` (or `*def`), only the scalar component of it.	[reply] [d/l] [select]
Re^3: Understanding typeglobs and local() by ysth (Canon) on Jan 13, 2006 at 04:10 UTC
x is a 'level higher' than \$x in the internal structure of the symbol table. Very close. x is a glob, a variable (storage location) of type PVGV. It knows what its own name and package are, and holds a value, a GP. The GP is the value that is copied with assignments like foo = bar. The GP holds all the various possible types (SV, AV, HV, CV, PVIO, etc.).	[reply]
Re^3: Understanding typeglobs and local() by dave_the_m (Monsignor) on Jan 13, 2006 at 12:25 UTC
From what I understand, local ($_); is roughly equivalent to: `my $saved = $_; $_ = undef; ... $_ = $saved;` [download] No, its actually equivalent to `my $tmp; my $old = \$_; _ = \$tmp; ... _ = \$old;` [download] as can be seen from the following: `$ perl -le '$x = 1; $r = \$x; local $x=2; print "$$r $x"' 1 2 $` [download] In your example they both would have printed 2. Dave.	[reply] [d/l] [select]