regex - 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. * --> *
2. ** --> **
3. ******* --> **
4. *abc**def* --> *abcdef*

The answer is:

str.replaceAll("(^\*)|(\*$)|\*", "$1$2");

I tried the answer on my machine and it works. But I don't know how it works.

From my understanding, all matched substrings should be replaced with $1$2. However, it works as:

1. (^\*) replaced with $1,
2. (\*$) replaced with $2,
3. \* replaced with empty.

Could someone explain how it works? More specifically, if there is | between expressions, how String.replaceAll() works with back reference?

Thank you in advance.

Answer 1

I'll try to explain what's happening in regex.

str.replaceAll("(^\*)|(\*$)|\*", "$1$2");

$1 represents first group which is (^\*) $2 represents 2nd group (\*$)

when you call str.replaceAll, you are essentially capturing both groups and everything else but when replacing, replace captured text with whatever got captured in both groups.

Example: *abc**def* --> *abcdef*

Regex is found string starting with *, it will put in $1 group, next it will keep looking until it find * at end of group and store it in #2. now when replacing it will eliminate all * except one stored in $1 or $2

For more information see Capture Groups