Ilya (>):
> look like bug (in r31017):
>
> my $k = 'bar';
> my %s;
> say defined $k; # 1
> say defined %s{$k}; # 0
> say defined $k and defined %s{$k}; # 1 (!) must be 0
> say 1 && 0; # 0
No, this is intended. "and" has very loose precedence, even looser
than the "say" call. The crucial statement is thus parsed like this:
(say defined $k) and defined %s{$k};
...which would output 1 but return 0.
To learn more about operator precedence, see
<http://perlcabal.org/syn/S03.html#Operator_precedence>
Also check out the excellent "perldoc perlop" for Perl 5 for more
examples of this.
<http://perldoc.perl.org/perlop.html>
// Carl
