in reply to Re^3: Why Perl boolean expression sometimes treated as lvalue?
in thread Why Perl boolean expression sometimes treated as lvalue?
If so, your message does not help. Please clarify the question if you think I didn't understand it.
Consider that you didn't miss the point -- in what way does your reply constitute an answer or an explanation?
foo( $blah && $foo ); works to alias $foo
$ref = \($blah && $foo); works to reference $foo
( $blah && $foo ) = foo(); is illegal
Where is this documented? Why is it illegal?
Oh, ikegami says it doesn't return a copy (why would it) -- who asked about a copy ?
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Re^5: Why Perl boolean expression sometimes treated as lvalue?
by ikegami (Patriarch) on Feb 11, 2013 at 21:58 UTC | |
by ikegami (Patriarch) on Feb 11, 2013 at 22:23 UTC |