Creating a fill-in regexp

[jpfarmer]

I'm working on a program where I'm dealing with long strings that look like this (only much longer): 0000010000000100000100001. I want to fill in the area between the 1s with zeros, so that string would become 0000011111111100000111111. I know that I can match the zeros with m/(?<=1)0+(?=1)/, but I don't know how to go about replacing all of the 0s with 1s just inside the match. Can I chain that with tr// somehow?

[shenme]

Use a 'substitute' ;-)

$s1 = '0000010000000100000100001';
$s1 =~ s/1(0+)1/'1'x((length($1)+2))/eg;

Output

 0000011111111100000111111

[jpfarmer]

It never occured to me that I could replace with a variable-length string. Oops. I feel pretty dumb now.

Thanks for your help!

[CombatSquirrel]

Use the s/// substitution operator:

   $string =~ s/(?<=1)(0+)(?=1)/'1' x length($1)/eg;

Hope this helped.
CombatSquirrel.

---

Entropy is the tendency of everything going to hell.

[Abigail-II]

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/

Abigail

[Enlil]

Here is one solution:

s/(10*1)/1 x length($1)/eg;

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

[Not_a_Number]

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