Java String.replaceAll() with back reference

Question

There is a Java Regex question: Given a string, if the \"*\" is at the start or the end of the string, keep it, otherwise, remove it. For example:

1. *

Answer 1

According to the Javadoc:

Note that backslashes () and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll. Use Matcher.quoteReplacement(java.lang.String) to suppress the special meaning of these characters, if desired.

Your regex: "(^\\*)|(\\*$)|\\*"

After removing quotes and String escapes: (^\*)|(\*$)|\*

There are three parts, separated by pipes |. The pipes mean OR, which means that replaceAll() replaces them with the stuff from the second part: $1$2. Essentially, the 1st part >> $1, the second >> $2, the third >> "". Note that "the 1st part" == $1, and so on... So it's not technically replaced.

1 (^\*) is a capture group (the first). ^ anchors to the string start. \* matches *, but needs the escape \.

2 (\*$) again, a capture group (2nd one). Difference here is it anchors to the end with $

3 \* like before, matches a literal *

The thing you need to understand about regexes is it will always take the first path if it matches. While *s at the beginning and end of the string could be matched by the 3rd part, they match the first or second parts instead.