The series “Yet Another Perl 6 Operator” is back with this brand new article

In the article on coercion operators, we got to know the prefix operator '?' which converts values into Bool::True or Bool::False. Like it happens with '~' for strings, '?' is recurrent for boolean operators.

In Perl 6, the usual infix boolean operators are: 

?& - and
?| - or
?^ - xor

These operators evaluate their operands in boolean context and apply simple Boole’s algebra on them. 

False ?& False  # False
''    ?& 'yes'  # False
1     ?& False  # False
42    ?& 42     # True

""    ?| 0      # False
False ?| True   # True
[1]   ?| 0      # True
True  ?| True   # True

''    ?^ ''     # False
undef ?^ {a=>1} # True
{:k}  ?^ undef  # True
True  ?^ True   # False

Each of the three operators always evaluate both sides and return one of the standard values Bool::True or Bool::False. So these boolean AND and OR do not short-circuit as their logical counterparts: '&&' and '||'. Precedence is different too. 

            Equivalent to         precedence

$a ?& $b    ?$a * ?$b != 0        multiplicative
$a ?| $b    ?$a + ?$b != 0        additive
$a ?^ $b    ?$a + ?$b == 1        additive

The boolean negation operator may be written '?^' or '!'. 

?^ $a       Equivalent to         True ?^  $a

Conceptually '?^' coerces to boolean first and then flips the bit. Synopsis 3 recommends the use of '!' instead.

Update: rhr at #perl6@freenode pointed how I mistakenly took [] and {} to be true in the examples above. In Perl 6, containers and hashes are true only if they have a least one member or pair, respectively. So I got it right.

After a too long pause, I am resuming the series to the delight of Perl 6 lovers and haters (though I don’t think they exist… the haters :).

Next article will be due… soon. (Not joking.)


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