StackOverflowError occured as using java.util.regex.Matcher

Question

I try to catch text by Regular Expression. I list codes as follows.

Pattern p=Pattern.compile("<@a>(?:.|\\s)+?</@a>"); 
Matcher m = p.matcher(fileContents.toString());
while(m.find()) {
    //Error will be thrown at this point
    System.out.println(m.group());
}

If the length of text I want to catch is too long, system will throw me a StackOverflowError. Otherwise, the codes work well. Please help me how to solve this problem.

How large is fileContents.toString()?

Bart Kiers
– Bart Kiers

2010-03-03 09:44:17 +00:00
Commented Mar 3, 2010 at 9:44 — Bart Kiers
– Bart Kiers, Commented Mar 3, 2010 at 9:44

Tim Pietzcker · Accepted Answer · 2010-03-03 09:38:15Z

3

The dot and \s both match whitespace characters. That might lead to unnecessary backtracking. What do you want to match? Probably any character, including linebreaks?

Then just use the lazy dot with the dot-matches-newlines option enabled:

Pattern p=Pattern.compile("<@a>.+?</@a>", Pattern.DOTALL);

You are aware that you'll run into trouble if <@a> tags can be nested in your input?

answered Mar 3, 2010 at 9:38

Tim Pietzcker

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2 Comments

Stephen C Over a year ago

In fact, the backtrack points are probably what is causing the stack overflow.

Tim Pietzcker Over a year ago

That's what I meant to imply (without having seen the text that is to be matched). Thanks for the clarification :)

Narayan Yerrabachu · Accepted Answer · 2013-05-22 06:52:39Z

-1

I always try to search text with case insensitive..
Pattern p=Pattern.compile("<@a>.+?</@a>", Pattern.DOTALL | Pattern.CASE_INSENSITIVE);

answered May 22, 2013 at 6:52

Narayan Yerrabachu

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