Re: Why Perl boolean expression sometimes treated as lvalue?

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

Quite simply:

$y always returns the scalar, not a copy of it.
&& always returns the scalar its LHS returned or the scalar(s) its RHS returned, not a copy of it.

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.

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Re^2: Why Perl boolean expression sometimes treated as lvalue? by Anonymous Monk on Feb 11, 2013 at 04:58 UTC
You're missing the point	[reply]
Re^3: Why Perl boolean expression sometimes treated as lvalue? by ikegami (Patriarch) on Feb 11, 2013 at 05:50 UTC
If so, your message does not help. Please clarify the question if you think I didn't understand it.	[reply]
Re^4: Why Perl boolean expression sometimes treated as lvalue? by Anonymous Monk on Feb 11, 2013 at 06:45 UTC
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 ?	[reply]
Re^5: Why Perl boolean expression sometimes treated as lvalue? by ikegami (Patriarch) on Feb 11, 2013 at 21:58 UTC
Re^6: Why Perl boolean expression sometimes treated as lvalue? by ikegami (Patriarch) on Feb 11, 2013 at 22:23 UTC

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