Re: Re: $1 trap

 sub chg {
       $_ [0] =~ $_ [1] && $1 ? "yes\n" : "no\n";
 }
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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/

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Re3: $1 trap by bbfu (Curate) on Aug 10, 2003 at 16:37 UTC
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]
Re: Re3: $1 trap by Chady (Priest) on Aug 11, 2003 at 05:43 UTC
NO. This would return "yes", because the regex will match, and thus will return true, even if there is no parens in the regex. That's why this works: `$_ = "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]
Re: Re: Re3: $1 trap by Crackers2 (Parson) on Aug 11, 2003 at 16:46 UTC
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.	[reply] [d/l] [select]