Re: Re: Golf this reg ex

japhy, thanks but I'm not sure I understand the /* and the fact that you can have, what appears to be, compounded reg exes. Anyway, I can't get it to work in

my $number = "..2314-1234123.";

if ($number =~ /\d/*/^([\d.-]+)$/ ) {
    print "$1\n";
}
[download]

If you have the time, a simple explanation would be educational. TIA

—Brad
"A little yeast leavens the whole dough."

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Re: Re: Re: Golf this reg ex by halley (Prior) on Apr 16, 2004 at 14:01 UTC
`perl -MO=Deparse` `my $number = '..2314-1234123.'; if ($number =~ /\d/ * /^([\d.-]+)$/) { print "$1\n"; }` [download] The original trick is multiplying the first regex result (as a scalar) by the second regex result (as a scalar). That is, 11 is true, but 01 is false. However, since the two regexen aren't being bound to `$number` in your example, the code actually doesn't work. The second regex is searching `$_` for matches instead. -- `[ e d @ h a l l e y . c c ]`	[reply] [d/l] [select]

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Re: Re: Re: Golf this reg ex
by halley (Prior) on Apr 16, 2004 at 14:01 UTC

perl -MO=Deparse

my $number = '..2314-1234123.';
if ($number =~ /\d/ * /^([\d.-]+)$/) {
    print "$1\n";
}
[download]

However, since the two regexen aren't being bound to $number in your example, the code actually doesn't work. The second regex is searching $_ for matches instead.

--
[ e d @ h a l l e y . c c ]

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