Typeglob syntax arcanery

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Here's what I want to do:

$pack1 = "alpha";
$sym1 = "beta";

$pack2 = "charlie";
$sym2 = "delta";

# insert magic here, that does this, without hardcoding the identifier
+s
# *alpha::beta = *charlie::delta
[download]

But I can't figure out the syntax. Help?

Comment on Typeglob syntax arcanery Download Code

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Re: Typeglob syntax arcanery by BrowserUk (Patriarch) on Mar 03, 2011 at 18:32 UTC
`sub charlie::delta{ print "from:sub charlie::delta" };; $pack1 = "alpha"; $sym1 = "beta"; $pack2 = "charlie"; $sym2 = "delta";; { "$pack1\::$sym1" } = { "$pack2\::$sym2" };; alpha::beta();; from:sub charlie::delta` [download] Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply] [d/l]
Re: Typeglob syntax arcanery by Fletch (Bishop) on Mar 03, 2011 at 18:29 UTC
Erm . . . `$ perl -le '$charlie::delta="foo";($a,$b,$c,$d)=qw( alpha beta charlie + delta ); {"${a}::$b"} = {"${c}::$d"};print $alpha::beta;' foo` [download] Wrap with a surrounding block of `no strict` as needed. Not that it's probably not a good idea to begin with. The cake is a lie. The cake is a lie. The cake is a lie.	[reply] [d/l] [select]
Re^2: Typeglob syntax arcanery by Anonymous Monk on Mar 03, 2011 at 18:34 UTC
Ahh, thanks. Perfect.	[reply]
Re: Typeglob syntax arcanery by Anonymous Monk on Mar 03, 2011 at 18:14 UTC
AND without using eval	[reply]