You don't need recursion. Just use the positive look back and look forward assertions which don't eat string. See perlman:perlre

my $a = "x|xx|xx|xxxx|xx|xxx|xx|xx|x";
$a =~ s/(?<=\|)xx(?=\|)/x x/g;
print "[$a]\n";
__DATA__
[x|x x|x x|xxxx|x x|xxx|x x|x x|x]
[download]

You don't say what you want to do with the edge cases:

$a = 'xx|xx|xx';
# should this be (use example above):
[xx|x x|xx]
# or should it be
[x x|x x|x x]
# in which case you will need to add an extra regex because you can't 
+have 
# variable width lookbacks. this regex just processes those edge cases
$a =~ s/^xx(?=\|)|(?<=\|)xx$/x x/g;
# combined with the first part gives you some real perl line noise
# if you want it all in a single regex
$a =~ s/^xx(?=\|)|(?<=\|)xx(?=\|)|(?<=\|)xx$/x x/g;
[download]

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

Hi Tachyon ,



      my $a = "x|xx|xx|xxxx|xx|xxx|xx|xx|x";
      $a =~ s/(?<=\|)xx(?=(\||$))/x x/g;

      # this is actually doing the regex edge cases too Tachyon
[download]

Well *actually* it does not. Your test string does not *actually* contain the edge cases ( xx at the begining and end of the string):o) If your test string had contained the edge cases you would no doubt have noted the following failure case.....

my $a = "xx|xx|xx|xxxx|xx|xxx|xx|xx|xx";
$a =~ s/(?<=\|)xx(?=(\||$))/x x/g;
print $a;
__DATA__
xx|x x|x x|xxxx|x x|xxx|x x|x x|x x
[download]

Here is one way to fix your regex:

$a =~ s/(?<=\|)|^xx(?=(\||$))/x x/g;

# and here is a way using the zero width boundary assertion \b
# that will match | ^ or $ but also matches any non aphlanumeric 
# so would potentially fail on '|xx,xx|' type strings
# if they exist in practice as your real data is no doubt
# not really xx.....
$a =~ s/\bxx\b/x x/g;
[download]

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

Hi Tachyon ,



     my $a = "x|xx|xx|xxxx|xx|xxx|xx|xx|x";
     $a =~ s/(?<=\|)xx(?=\|)/x x/g;
     print "[$a]\n";
[download]



This is the backward and forward assertions that I needed ,man , I have to learn more on this ,man THANKS A BUNCH Tachyonfor the help

Glad to help. The only bummer is that you cna't have variable width lookbacks (positive or negative) like (?<=x*) or (?<!x*) You can have variable width look forward assertions though (?=x*) or (?!x*) ar OK.....

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

See how you get on with this?

$s = "x|xx|xx|xxxx|xx|xxx|xx|xx|x";
$s =~ s[(x\|x)(x\|x)([x\|]*)][$1 $2$3]g;
print $s;
x|x x|xx|xxxx|xx|xxx|xx|xx|x
[download]

Hi BrowserUk,



My request was not clear, I think , Actually I need all the |xx| changed to |x x| and hence now I have used the tachyon answer of backward and forward assertion which I did not know much on before and I have seen your node once that uses < > code and then assigning $1 to a local variable inside the regex itself which was a lovely piece of regex , but I sure have to read and practice it thoroguhly BrowserUk, about backward, forward asssertions,and few other concepts of regex,THANKS A BUNCH

Sorry. I misunderstood your requirement (though it could have been stated a little more clearly:^).

This one (I think does what you want, edge case as well and is relatively simple.

$s = "xx|xx|xxxx|xx|xxx|xx|xx|xx";
$s =~ s[(?<!x)(x)(x)(?!x)][$1 $2]g;
print $s;
x x|x x|xxxx|x x|xxx|x x|x x|x x
[download]

Update Thinking about it, stick with tachyon's as this will fall over if your data contains any occurance of xx without |'s or x's on either side.

---

Examine what is said, not who speaks.

The 7th Rule of perl club is -- pearl clubs are easily damaged. Use a diamond club instead.

my $a = "x|xx|xx|xxxx|xx|xxx|xx|xx|x";
1 while ($a =~ s/\|xx\|/|x x|/);
print "[$a]\n";
__DATA__
[x|x x|x x|xxxx|x x|xxx|x x|x x|x]
[download]

but as tachyon mentioned: what to do with the edge cases? (this method is not optimal by any means.)

-enlil