No. && has higher precedence than ||, so this parses as (1 || (0 && 0)), which is equivalent to (1 || 0).
See http://www.perl.com/doc/manual/html/pod/perlop.html Regards, Jason Elbaum On 2/8/07, Yossi Itzkovich <[EMAIL PROTECTED]> wrote: > Hi, > > For this code: > perl -we '$a=(1||0&&0); print $a' > I get "1", although I should get "0" . > > Isn't it a bug ? > > I am using perl v5.8.4 > > Thanks > > Yossi > > _______________________________________________ > Perl mailing list > [email protected] > http://perl.org.il/mailman/listinfo/perl > _______________________________________________ Perl mailing list [email protected] http://perl.org.il/mailman/listinfo/perl
