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Perl regular expression question help

Posted on 2003-03-24
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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?  
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Question by:regulatorz
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Expert Comment

by:prady_21
ID: 8200467
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
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Author Comment

by:regulatorz
ID: 8200816
Ok so for this one say if $0 has "/test/blah.pl" then $1 will have "/test" or empty?
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by:prady_21
ID: 8201966
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 earned 80 total points
ID: 8202007
The first regular expression is searching $0 for:
    Any number of any characters, then
    /, then
    one or more characters excluding slashes
If this is found then the first character string is stored in $1 and is added to the front of the @INC array.

The first regular expression is similar, except it uses backslashes instead of slashes.

To observe the operation, run this script:

($0=~ m,(.*)/[^/]+,)   && unshift (@INC, "$1");
($0=~ m,(.*)\\[^\\]+,) && unshift (@INC, "$1");
print join "\n", @INC;
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by:holli
ID: 8205906
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

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Author Comment

by:regulatorz
ID: 8207009
thanks for all the help.  
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