Isn't javascript regex lazy?

Question

Lets say I have the input value:

var input_value = "** foo ** bar **";

Then I run this:

input_value = input_value.replace(/(\*\*)([\S\s.]+)(\*\*)/g,replacer_markup);

function replacer_markup(match, p1, p2, p3, offset, string) {
  var pre = '';
  var post = '';
  if (p1 == '**')
  {
    pre = '<b>';
    post = '</b>';
  }
  return pre+p2+post
}

It will give me:

foo ** bar

But instead I want to have:

foo bar **

I want to have the smallest possible result. How can I achieve this?

Sorry for any spelling mistakes, I'm not a native speaker. Thank you for your time.

hwnd · Accepted Answer · 2015-04-07 01:00:09Z

3

+ is a greedy operator — it tells the engine to match as much as it can and still allows the remainder of the regular expression to match. Use +? for a non-greedy match meaning "one or more — preferably as few as possible". Also, you can exclude the . from the character class.

([\S\s]+?)

edited Apr 7, 2015 at 1:00

answered Apr 7, 2015 at 0:50

hwnd

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elclanrs · Accepted Answer · 2015-04-07 00:50:15Z

2

Make the regex non-greedy:

/(\*\*)([\S\s.]+?)(\*\*)/g
               -^-

answered Apr 7, 2015 at 0:50

elclanrs

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