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.

In reply to Re^2: Understanding typeglobs and local() by whio
in thread Understanding typeglobs and local() by whio

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