Re: Creating a fill-in regexp

[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!