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Perl regular expression question help

Ok I'm new programmer to Perl and all that regular expressin stuff isn't going well with me. Basically I'm trying to understand the following 2 lines of code:

($0=~ m,(.*)/[^/]+,) && unshift (@INC, "$1");
($0=~ m,(.*)\\[^\\]+,) && unshift (@INC, "$1");

Can someone please explain what that regular expression is comparing to? Is that m there the match keyword? Isnt the format for match m//? I'm confused! What is been put in the array @INC?

prady_21

First of all
m// is the matching operator as you know, but perl allows any kind of specifiers, for ex m, , is also a matching specifier, and also m. .
http://webdevelopment.developersnetwork.com/Articles.asp?Article=143

And for your next question, when we do matching like
m// or m,, or anything of that sort, then anything that is specified in the brackets ie () is assigned the value $1.
ie m/(.*)/ puts the value that matches (.*) into $1.
see ->
http://www.comp.leeds.ac.uk/Perl/sandtr.html

regulatorz

ASKER

Ok so for this one say if $0 has "/test/blah.pl" then $1 will have "/test" or empty?

prady_21

The first expr matches /test ($1 = "/test"
and i think the second expr is used for filenames of the type \test\blah.pl

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PC_User321

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holli

some notes on regex.
once you see .* in a regex, hold on and think how you can avoid it. that is because .* can very easily cause heavy backtracking what makes your regex slow.

a better way (and a non-platform-specific one) for your code above is this:

BEGIN { use File::Basename; unshift ( @INC, (fileparse($0))[1] ); }

holli

regulatorz

ASKER

thanks for all the help.