why didn't I get an error from perl when $v=false;

Question

On commandline I wrote:

perl -e '$v=false; if($v==true){print 1;}'

and instead of producing an error it returned "1"!

As I remember in past this returned an error to remind me that perl has no boolean type. anything changed in perl in the meantime? or did I remember incorrectly and this always was valid code?

https://perlmaven.com/barewords-in-perl. Always `use strict` and `use warnings`. — ggorlen, Aug 10 '20 at 23:00
many thanks! I edited .bashrc and added `alias perl="perl -Mstrict -Mwarnings"` — user11566470, Aug 10 '20 at 23:16
Instead of aliasing `perl`, you can set `PERL5OPT` with the default options. — brian d foy, Aug 11 '20 at 17:10

score 8 · Accepted Answer · answered Aug 10 '20 at 23:03

Without strict, the bareword false is interpreted as the string ('false'). Using the numeric == numifies it to 0, and "true" as a number is zero, too.

B::Deparse shows it clearly:

$ perl -MO=Deparse -e '$v=false; if($v==true){print 1;}'
$v = 'false';
if ($v == 'true') {
    print 1;
}

warnings would have warned you 4 times: twice about the unquoted bareword, and twice about numeric comparison of strings.

why didn't I get an error from perl when $v=false;

1 Answers1