Re: Creating a fill-in regexp

Use a 'substitute' ;-)

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

Output

 0000011111111100000111111

Comment on Re: Creating a fill-in regexp Download Code

Replies are listed 'Best First'.
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]