Wouldn't this transpose the second word and the first word also?

-----Original Message-----
From: Nikola Janceski [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 24, 2002 1:25 PM
To: 'Naveen Parmar'; [EMAIL PROTECTED]
Subject: RE: Pattern matching


[^ ]+ (means match one or more charaters that is not a space (literal " ") 
s/^([^ ]+) +([^ ]+)/$2 $1/; reads

s/ Substitute,
^  that which begins the string with
([^ ]+) one or more charaters that is not a space (store in $1)
 + one or more charaters that IS a space
([^ ]+) one or more charaters that is not a space (store in $2)
/ with
$2 $1
/ end Substitute;

so that would take:

$string = "dsljfsldj     lsjdkfldjlf     lsdkjflkdjssdlf jslj lk sf";
$string =~ s/^([^ ]+) +([^ ]+)/$2 $1/;

$string is now "dsljfsldj lsjdkfldjlf     lsdkjflkdjssdlf jslj lk sf"

nothing would happen if string started with a space.


-----Original Message-----
From: Naveen Parmar [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 24, 2002 4:16 PM
To: [EMAIL PROTECTED]
Subject: Pattern matching


What is the purpose of using [^"] in the following expression:

s/^([^ ]+) +([^ ]+)/$2 $1/;



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