regex and multiplication

elwarren has asked for the wisdom of the Perl Monks concerning the following question:

Simple question. Can I execute on the right side of a regex? I'm searching for a word, and I want to double whatever number comes after it. This doesn't do what I want:

$var =~ s/WORD\s(\d+)/WORD (2 * $1)/;
[download]

I know I can do it with an if statement and $1, but now that I've started down this path...

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(Ovid) Re: regex and multiplication by Ovid (Cardinal) on Sep 21, 2000 at 20:19 UTC
Yes, add the /e switch: `$var =~ s/WORD\s(\d+)/"WORD " . (2 * $1)/e;` [download] See perlre for details. Cheers, Ovid Update: Fixed post in response to tye's clarification. Join the Perlmonks Setiathome Group or just go the the link and check out our stats.	[reply] [d/l]
RE: Re: regex and multiplication by tye (Sage) on Sep 21, 2000 at 20:27 UTC
You should quote the parts of the replacement you don't want executed: `s/WORD\s(\d+)/"WORD ".(2$1)/e;` [download] Or, how I'd write it: `s/(WORD\s)(\d+)/$1.(2$2)/e;` [download] - tye (but my friends call me "Tye")	[reply] [d/l] [select]
Re: regex and multiplication by elwarren (Priest) on Sep 21, 2000 at 20:26 UTC
doh! I knew it was something easy like that. I don't see it anywhere on the perlre page, because I must have gone over it three times this morning. As soon as I read your reply I found it in the perlop page. Thanks, good response time!	[reply]