NO. This would return "no", because he's not only comparing the
result of the regex, but also the value of the submatch:

"w0rd" =~ /(\d)/ && $1 ? print"Yes": print"no";

The regex matches, so $1 becomes 0. This makes the condition:

(true) && 0 ? print"Yes": print"no";

Which evaluates to false.

In reply to Re: Re: Re3: $1 trap by Crackers2
in thread $1 trap by chunlou

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