Re^7: 'xor' operator is not a sibling to 'or' and 'and'?

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])
[download]

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], ["", ""])
[download]

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery
FootballPerl is like chess, only without the dice}

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