in reply to Re^5: 'xor' operator is not a sibling to 'or' and 'and'?
in thread 'xor' operator is not a sibling to 'or' and 'and'?

>> A xor B := ( A and not B) or ( B and not A)

I only want to leave here an example: although 'logically' they return the same 'Perl-truthness', in this case they won't return same value: 
perl -wle 'my( $A, $B ); print for map "[$_]", ( $A xor $B ), ( $A and
+ not $B or $B and not $A ); '

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output ('' vs. undefined): 
Use of uninitialized value $_ in concatenation (.) or string at -e lin
+e 1.
[]
[]

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Re^7: 'xor' operator is not a sibling to 'or' and 'and'?
by LanX (Saint) on Dec 28, 2019 at 17:06 UTC
    yep 
    perl -MData::Dump=pp
@values=("true",undef);
for $A (@values) {
    for $B (@values) {
        pp [$A,$B], [ ($A xor $B), ( ($A && !$B) || ($B && !$A) ) ];
        }
}
__END__
(["true", "true"], ["", ""])
(["true", undef], [1, 1])
([undef, "true"], [1, 1])
([undef, undef], ["", undef])

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    but if you apply De Morgan's law, you'll always get "default" Booleans:

    perl -MData::Dump=pp
@values=("true",undef);
for $A (@values) {
    for $B (@values) {
        pp [$A,$B], [ ($A xor $B), !( ( !$A || $B) && (!$B || $A) ) ];
    }
}
__END__
(["true", "true"], ["", ""])
(["true", undef], [1, 1])
([undef, "true"], [1, 1])
([undef, undef], ["", ""])

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    Cheers Rolf
    (addicted to the Perl Programming Language :)
    Wikisyntax for the Monastery FootballPerl is like chess, only without the dice

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