in reply to $1 trap

This is a well-known "trap", and can easily happen without creating such a sub. The problem is your assumption that a failed regular expression would somehow undefine $1. But it doesn't. The proper way of writing your sub is: 
    sub chg {
        $_ [0] =~ $_ [1] && $1 ? "yes\n" : "no\n";
    }

[download]

Abigail

Replies are listed 'Best First'.
Re: Re: $1 trap
by Chady (Priest) on Aug 10, 2003 at 16:07 UTC 
     sub chg {
       $_ [0] =~ $_ [1] && $1 ? "yes\n" : "no\n";
 }

    [download]

    Isn't the && $1 redundant here? or am I missing something out?

    He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life.

    Chady | http://chady.net/
[reply]
[d/l]
[select]

      The OP's code was actually returning "yes" if $1 contained a true value. So, the following would've returned "no":

      chg("w0rd", qr/(\d)/); # that's a zero, not capital O

      bbfu
      Black flowers blossom
      Fearless on my breath

[reply]
[d/l]
        NO.

        This would return "yes", because the regex will match, and thus will return true, even if there is no parens in the regex.

        That's why this works: 
        $_ = "something blah blah";
if (/something/) {
     print "Ok";
}

        [download]
        He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life.

        Chady | http://chady.net/
[reply]
[d/l]

In Section Seekers of Perl Wisdom