Re^5: Burned by precedence rules (== true)

[ikegami]

in reply to Re^4: Burned by precedence rules (== true)
in thread Burned by precedence rules

There's a logical operator for that: xor

if ( !$x == !$y) ) # bool($x) == bool($y)
if ( !$x != !$y  ) # bool($x) != bool($y)

can be written as

if ( !($x xor $y) ) # bool($x) == bool($y)
if (   $x xor $y  ) # bool($x) != bool($y)

Unforunately, both sets have readability issues.

[TimToady]

In Perl 6 we can use positive boolean context, which helps readability a little:

if ?$x == ?$y {...}
if ?$x != ?$y {...}

We can use junctions, which again helps the different case more than the same case:

if not $x ^ $y {...}
if $x ^ $y {...}

or if you prefer:

if !one($x,$y) {...}
if one($x,$y) {...}

That's probably enough ways to do it...

[tye]

Nah. Clearly this demonstrates the utility of a few "Boolean comparison" operators:

    if $isFree ?= $isEasy
    if $isOne !?= $isOther

        unlink $dev
    eqv unlink $prod
    or  die "Inconsistent state";

         system $verifyUser
    neqv system $verifyAdmin
    or   die "You are not just either a User or an Admin";

;)

[TimToady]

Well, we can't use eqv because we already use that for something else, but we do have a Boolean not-equal already:

    $x ?^ $y

What we don't have is a Boolean equal. I thought about adding:

    $x ?= $y

but people would probably think that means:

    $x = $x ? $y;

which is nonsense, but they'd still think it. :-)

Anyway, the utility is not worth the extra complexity, in my opinion. Comparing two Booleans for equivalence is an odd enough thing to do that I don't mind if it looks a little odd.

[demerphq]

Ah yes, scalar xor, i always forget about it. Personally the !$x == !$y makes more sense to me than !($x xor $y). But whatever.