I think the rest after the 'if' for the last is wrong. either do this: last if ($counter > 2);
or if ( $counter > 2) { print 'if : ' . $counter . "\n"; #could do print "if : $counter\n" as well last; } On Mon, Jun 17, 2013 at 8:56 AM, lee <l...@yun.yagibdah.de> wrote: > Hi, > > trying to figure out what `last' actually does, I wrote this test > script: > > > use strict; > use warnings; > use autodie; > > my $counter = 0; > > while($counter < 8) { > last if($counter > 2) { > print "if: " . $counter . "\n"; > } > else { > print "else: " . $counter . "\n"; > } > $counter++; > } > > > Unfortunately, that gives syntax errors. I would expect the following > output: > > > else: 0 > else: 1 > else: 2 > if: 3 > > > What's wrong with that (letting aside that it is terribly ambiguous)? > > > -- > "Object-oriented programming languages aren't completely convinced that > you should be allowed to do anything with functions." > http://www.joelonsoftware.com/items/2006/08/01.html > > -- > To unsubscribe, e-mail: beginners-unsubscr...@perl.org > For additional commands, e-mail: beginners-h...@perl.org > http://learn.perl.org/ > > > -- wapembe seller of truths, half and whole inquire about quantity discounts!