Hi Shawn! On Sat, 1 Jul 2017 11:32:30 -0400 Shawn H Corey <shawnhco...@gmail.com> wrote:
> On Sat, 1 Jul 2017 17:27:02 +0200 > hw <h...@gc-24.de> wrote: > > > > > Hi, > > > > can someone please explain this: > > > > > > perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; print "i: $i\n";' > > i: > > > > > > Particularly: > > > > > > + Why doesn´t it print 1? > > Because !!$i is zero > > > > > + How is this not a bug? > > Nope, no bug. > > > > > + What is being printed here? > > !!$i which is !(!(0)) which is !(1) which is 0 > I suspect !1 returns an empty string in scalar context. > > > > + How do you do what I intended in perl? > > > > How do we know what you intend? > > -- ----------------------------------------------------------------- Shlomi Fish http://www.shlomifish.org/ http://shlomifishswiki.branchable.com/Self-Sufficiency/ I don’t believe in fairies. Oops! A fairy died. I don’t believe in fairies. Oops! Another fairy died. — http://www.shlomifish.org/humour.html -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
