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.

In reply to Re: Why Perl boolean expression sometimes treated as lvalue? by ikegami
in thread Why Perl boolean expression sometimes treated as lvalue? by vsespb

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