http://www.perlmonks.org?node_id=1018092


in reply to Why Perl boolean expression sometimes treated as lvalue?

Quite simply:

That's it. Why isn't it documented that they don't make a copy in rvalue context? Why would it.

If you wanted to explicitly copy the scalar, you could do a( 0+( $x && $y ) ). Then, $_[0] = 3; will modify the anonymous scalar the addition constructed.