Re^4: Perl RE; how to capture, and replace based on a block?

[taint]

Greetings, Laurent_R, and thanks for the reply.

Oh yes. I'm keen on the \n v \r v \n\r thing, and you're absolutely correct. Except, in my case, I'm on a *NIX box, and I've written the files myself. So I know they're utf-8 (no BOM), with newlines, no "hard" returns. :)

Maybe it's just the examples I was reading (perlrequick, perlretut, and perlfaq6) but I got the impression that Perl RE wasn't greedy. More Perl RE reading, I guess.

Thanks again, for the response Laurent_R.

--Chris

[Laurent_R]

Hi Chris

just a quick example on greedy and non greedy quantifiers, under the Perl debugger:

  DB<1> $string = "foo bar foo bar foo";

  DB<2> x $string =~ /(fo.*ar)/
0  'foo bar foo bar'
  DB<3> x $string =~ /(fo.*?ar)/
0  'foo bar'
  DB<4>

As you can see the default "*" quantifier is greedy, so that the regex tries to match as much as possible. Using the modified "*?" quantifier, the regex becomes non greedy. You will see the same type of result if you try the "+" and the "+?" quantifiers.

[taint]

Thanks Laurent_R for the help.

Your example helps better explain why I got the impression Perl was not greedy, the way sed is. Most all the examples I read in the perldocs, used .*?. But my experiments, and re-training myself with Perl RE's. Seems to show, once I get out of the sed habits. Perl will be a lot more flexible, and powerful. I should have started this a l-o-o-n-g time ago. :P

Thanks again, Laurent_R. I really appreciate it.

--Chris