Creating a fill-in regexp

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

Replies are listed 'Best First'.
Re: Creating a fill-in regexp by shenme (Priest) on Sep 24, 2003 at 18:50 UTC
Use a 'substitute' ;-) `$s1 = '0000010000000100000100001'; $s1 =~ s/1(0+)1/'1'x((length($1)+2))/eg;` [download] Output 0000011111111100000111111	[reply] [d/l]
Re: Re: Creating a fill-in regexp by jpfarmer (Pilgrim) on Sep 24, 2003 at 19:15 UTC
It never occured to me that I could replace with a variable-length string. Oops. I feel pretty dumb now. Thanks for your help!	[reply]
Re: Creating a fill-in regexp by Not_a_Number (Prior) on Sep 24, 2003 at 19:34 UTC
As a sort of follow-up to Abigail-II's comment, note that for the string: `101110101` the three snippets posted so far give three different 'solutions' :-) dave	[reply] [d/l]
Re: Creating a fill-in regexp by CombatSquirrel (Hermit) on Sep 24, 2003 at 18:51 UTC
Use the `s///` substitution operator: `$string =~ s/(?<=1)(0+)(?=1)/'1' x length($1)/eg;` [download] Hope this helped. CombatSquirrel. Entropy is the tendency of everything going to hell.	[reply] [d/l] [select]
Re: Creating a fill-in regexp by Abigail-II (Bishop) on Sep 24, 2003 at 18:53 UTC
Your description doesn't match your example. You say you want to fill the area between the 1's with zero's, but in your example, there are three area's between 1's, and out of the three, you replace two of them with 1's. If your question is, "how do I replace the 0's with 1's", use tr: `tr/0/1/` [download] Abigail	[reply] [d/l]
Re: Creating a fill-in regexp by Enlil (Parson) on Sep 24, 2003 at 18:55 UTC
Here is one solution: `s/(101)/1 x length($1)/eg;` [download] Note that you might want to change the to a + depending on what you want to happen should the ones be right next to each other. updatewording was "change the * to a star", thanks L~R -enlil	[reply] [d/l]