Confusing behavior of a capturing group in a positive lookbehind in a Java regex with Pattern.matcher - Stack Overflow

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The following issue is observed only on Java and not on other regex flavors (e.g PCRE).

I have the following regex: (?:(?<=([A-Za-z\d]))|\b)(MyString). There's a capturing group on [A-Za-z\d] in the lookbehind.

And I'm trying to match (through Pattern.matcher(regex); to be precise, I'm calling replaceAll) the following string: string.MyString.

On PCRE, I will match MyString, and it will be the second group in the match. On Java, however, I will match the g in string as group 1, and MyString as group 2.

1. Why does Java do that? To me this regex implies that a character matching [A-Za-z\d] should only be matched if it directly precedes MyString, which is not the case here.
2. How can I avoid that and not match this g? I want to keep the capturing group in case I have to match a string like stringMyString, in which case I do need that g as group 1.

The following issue is observed only on Java and not on other regex flavors (e.g PCRE).

I have the following regex: (?:(?<=([A-Za-z\d]))|\b)(MyString). There's a capturing group on [A-Za-z\d] in the lookbehind.

And I'm trying to match (through Pattern.matcher(regex); to be precise, I'm calling replaceAll) the following string: string.MyString.

On PCRE, I will match MyString, and it will be the second group in the match. On Java, however, I will match the g in string as group 1, and MyString as group 2.

1. Why does Java do that? To me this regex implies that a character matching [A-Za-z\d] should only be matched if it directly precedes MyString, which is not the case here.
2. How can I avoid that and not match this g? I want to keep the capturing group in case I have to match a string like stringMyString, in which case I do need that g as group 1.

• java
• regex
• regex-lookarounds

Share Improve this question Follow asked Mar 28 at 11:40 wouldnotliketowouldnotliketo 33522 silver badges1414 bronze badges 1

• Looks like the Java regex engine does not reset Group 1 contents upon a failed match and once the match is found, the submatch is returned with the match. Looks like a bug to me, but it is probably related to regex specific Java functions. – Wiktor Stribiżew Commented Mar 28 at 11:54

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1 Answer 1

Sorted by: Reset to default 5

There is a line on the java.util.regex.Pattern docs

The captured input associated with a group is always the subsequence that the group most recently matched. If a group is evaluated a second time because of quantification then its previously-captured value, if any, will be retained if the second evaluation fails. Matching the string "aba" against the expression (a(b)?)+, for example, leaves group two set to "b". All captured input is discarded at the beginning of each match.

I think this line explains the behavior:

• If a group is evaluated a second time because of quantification then its previously-captured value, if any, will be retained if the second evaluation fails.

The last line:

• All captured input is discarded at the beginning of each match.

So if you have this string:

string.MyString.srting.MyString

And this regex:

(?:(?<=([tr]))|\b)(MyString)

You can see that the group 1 value is different in both matches as all captured input is discarded.

See an example on regex101

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