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Understanding possessive quantifiers, java regex

开发者 https://www.devze.com 2023-03-15 17:50 出处:网络
I understand that a possesive regex would go to the end of the text and would not backtrack to see if there was a match before the end. If at the end there\'s a match it returns true, otherwise it imm

I understand that a possesive regex would go to the end of the text and would not backtrack to see if there was a match before the end. If at the end there's a match it returns true, otherwise it immidiatly returns false. I've tride this:

Pattern patt = Pattern.compile(".*+foo");
Matcher matcher = patt.matcher("xxfooxxxxxfooxxxfoo");
while (matcher.find())
    System.out.println(matcher.group());

It giv开发者_JAVA百科es me nothing even though there's a match in the end. Any ideas why?

Also I understand that to make a regex lazy/possessive I add ?/+ after the first quantifier (i.e. *? or *+). Is that right? Thanks!


It gives me nothing even though there's a match in the end. Any ideas why?

The .*+ will match the entire input string (including the last foo). And because it does not backtrack from the end of the string, the regex .*+foo does not match.

Also I understand that to make a regex lazy/possessive I add ?/+ after the first quantifier (i.e. *? or *+). Is that right?

The counter part of possessive is not lazy. That would be greedy, which * by default is.

So, the regex .*?foo would match "xxfoo" and the regex .*foo would match "xxfooxxxxxfooxxxfoo".


Possessive quantifiers will not give up matches on a backtrack. The .*+ matches your entire string and then there's nothing for foo to match.

Er, like Bart said. :)

Use possessive quantifiers only when you know that what you've matched should never be backtracked (e.g., [^f]*+.*foo or, if you know that the only "f" characters will be at the start of "foo", [^f]*+foo).

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