Could you explain some of these many other ways to create aliases that end up in something different?

I named other ways in this node: @_, foreach, map, grep.

  DB<76> sub change { $_[0] = 666 }

  DB<77> $a=42

  DB<78> change($a)

  DB<79> p $a
666
  DB<80> change(42)
Modification of a read-only value attempted at (eval 86)[c:/Perl_524/l
+ib/perl5db.pl:737] line 2.

  DB<81>
[download]

still "lame"?

update

BUT!!!

  DB<81> sub change_arr { $_[0] = [666] }

  DB<82> $a=[42]

  DB<83> change_arr($a)

  DB<84> x $a
0  ARRAY(0x3363180)
   0  666
  DB<85> change_arr([42]) #NO ERROR???

  DB<86>
[download]

This is no answer to my question. My point is: however created, an alias is a reference to something put into a slot of a symbol table entry. I do not see any evidence in your examples against this claim.

UPDATE And then again, a de-referenced reference should behave similar to an alias. AFAICS it does.

Greetings,
-jo

$gryYup$d0ylprbpriprrYpkJl2xyl~rzg??P~5lp2hyl0p$

> put into a slot of a symbol table entry.

An alias is always a variable true, that's an obvious no-brainer.

But typeglob mechanisms are only a part of the way to create them.

Especially in for my $a (@a) { ... } there is no symbol table entry like you claimed. my $a is a private variable and an alias.

And newer Perl versions allow activating the feature for refaliasing too, again no symbol table.

Cheers Rolf
_{(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery}