> @$foo{bar}
What you intend by the above is the scalar value
in $foo{bar} dereferenced as an array by the @.
But that's not what happens. To achieve what you
want, use:
@{$foo{bar}}
The issue here is precedence of sigils (@, $, etc.)
versus subscript parens ({}, []), and the meaning
of various combinations of sigils and subscripts.
Sigils have a higher precedence than subscripts.
To force the meaning you wanted, we, well,
forced the meaning you wanted.
Fwiw, @$foo{bar} specifies that you want a list,
consisting of the values corresponding to the list
of keys in the parens (in this case a list of length
1, namely bar), looked up in the hash referenced
by $foo.
So, this should work:
%h = (key=>'value', key2=>'value2');
$foo = \%h;
print @$foo{key, key2};
hth.
