Question

The following Java regex “seems” to be working . The intent is to remove the escapeChar – backslash "\". That is "\\{" should become "{".

My question is

1. Isn’t the 10 char in the regex field – the closing parenthesis “)” – closing the regex group that began at char5? So how is this working for the chars after the closing parenthesis at char10?

2. Can someone break this regex down for me?

   str = str.replaceAll("\\\\([{}()\\[\\]\\\\!&:^~-])", "$1");
   

Answer 1

Isn’t the 10 char in the regex field – the closing parenthesis “)” – closing the regex group that began at char5? So how is this working for the chars after the closing parenthesis at char10?

No. The parentheses, both ( and ) are not meta-characters inside a character class. Note that inside a character class only these characters ^-[]\ have special meaning.

In the case of the caret (^) and the dash (-) they lose their special meaning if placed strategically within the char class: the caret if it’s placed anywhere but the beginning, and the - if it’s placed in the beginning or the end.

Can someone break this regex down for me?

Let’s remove the double escapes needed by Java, which turns \\\\([{}()\\[\\]\\\\!&:^~-]) into:

\\([{}()\[\]\\!&:^~-])   # the actual regex

Which breaks down into:

\\                   # match literal backslash
 (                   # open capture group
  [                  # open character class, matching any of
   {}()\[\]\\!&:^~-  # these characters: {}()[]\!&:^~-
  ]                  # close character class
 )                   # close capture group

Basically it says: match a backslash, followed by one of these characters {}()[]\!&:^~-, and put it into a capture group. This capture group is used in the replacement ($1), which replaces the whole match (backlash + character) with the character itself.

In other words, this removes leading backslashes from those special characters.