Java regex escaped characters

Question

When matching certain characters (such as line feed), you can use the regex \"\\\\n\" or indeed just \"\\n\". For example, the following splits a string into an array of lines:<

Answer 1

Yes there are different. The Java Compiler has different behavior for Unicode Escapes in the Java Book The Java Language Specification section 3.3;

The Java programming language specifies a standard way of transforming a program written in Unicode into ASCII that changes a program into a form that can be processed by ASCII-based tools. The transformation involves converting any Unicode escapes in the source text of the program to ASCII by adding an extra u - for example, \uxxxx becomes \uuxxxx - while simultaneously converting non- ASCII characters in the source text to Unicode escapes containing a single u each.

So how this affect the /n vs //n in the Java Doc:

It is therefore necessary to double backslashes in string literals that represent regular expressions to protect them from interpretation by the Java bytecode compiler.

An a example of the same doc:

The string literal "\b", for example, matches a single backspace character when interpreted as a regular expression, while "\b" matches a word boundary. The string literal "(hello)" is illegal and leads to a compile-time error; in order to match the string (hello) the string literal "\(hello\)" must be used.