Re^5: Perl RE; how to capture, and replace based on a block?


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Re^5: Perl RE; how to capture, and replace based on a block?

by Laurent_R (Canon)

on Dec 18, 2013 at 18:36 UTC ( [id://1067674]=note: print w/replies, xml )

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

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>
[download]

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.

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Re^6: Perl RE; how to capture, and replace based on a block?
by taint (Chaplain) on Dec 18, 2013 at 20:18 UTC

Laurent_R

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