title: format hex string using replace, regex and notepad++
body: <p>Given an hex string, which is unformatted, e.g.</p>
  
  <blockquote>
    <p>00,01,02,03,04,05,06,07,08,09,0a,0b,0c,0d,0e,0f,10,11,12,13,14,15,16,17,18,19,1a,1b,1c,1d,1e,1f,20,21,22,23,24,25,26,27,28,29,2a,2b,2c,2d,2e,2f,30,31,32,33,34,35,36,37,38,39,3a,3b,3c,3d,3e,3f,40,41,42,43,44,45,46,47,48,49,4a,4b,4c,4d,4e,4f,50,51,52,53,54,55,56,57,58,59,5a,5b,5c,5d,5e,5f,</p>
  </blockquote>
  
  <p>And a matching find regex for the first 23 hex values:</p>
  
  <pre><code>([[:xdigit:]][[:xdigit:]],){23}
  </code></pre>
  
  <p>How should the replace string look like,  that after the matching pattern an \crlf (\\r\n) is added to the string so that the block would looks like</p>
  
  <pre><code>00,01,02,03,04,05,06,07,08,09,0a,0b,0c,0d,0e,0f,10,11,12,13,14,15,16,
  17,18,19,1a,1b,1c,1d,1e,1f,20,21,22,23,24,25,26,27,28,29,2a,2b,2c,2d,
  2e,2f,30,31,32,33,34,35,36,37,38,39,3a,3b,3c,3d,3e,3f,40,41,42,43,44,
  45,46,47,48,49,4a,4b,4c,4d,4e,4f,50,51,52,53,54,55,56,57,58,59,5a,5b,
  5c,5d,5e,5f,
  </code></pre>