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

There is no difference in the current scenario. The usual string escape sequences are formed with the help of a single backslash and then a valid escape char ("\n", "\r", etc.) and regex escape sequences are formed with the help of a literal backslash (that is, a double backslash in the Java string literal) and a valid regex escape char ("\\n", "\\d", etc.).

"\n" (an escape sequence) is a literal LF (newline) and "\\n" is a regex escape sequence that matches an LF symbol.

"\r" (an escape sequence) is a literal CR (carriage return) and "\\r" is a regex escape sequence that matches an CR symbol.

"\t" (an escape sequence) is a literal tab symbol and "\\t" is a regex escape sequence that matches a tab symbol.

See the list in the Java regex docs for the supported list of regex escapes.

However, if you use a Pattern.COMMENTS flag (used to introduce comments and format a pattern nicely, making the regex engine ignore all unescaped whitespace in the pattern), you will need to either use "\\n" or "\\\n" to define a newline (LF) in the Java string literal and "\\r" or "\\\r" to define a carriage return (CR).

See a Java test:

String s = "\n";
System.out.println(s.replaceAll("\n", "LF")); // => LF
System.out.println(s.replaceAll("\\n", "LF")); // => LF
System.out.println(s.replaceAll("(?x)\\n", "LF")); // => LF
System.out.println(s.replaceAll("(?x)\\\n", "LF")); // => LF
System.out.println(s.replaceAll("(?x)\n", "")); 
// => 
//

Why is the last one producing +newline+? Because "(?x)\n" is equal to "", an empty pattern, and it matches an empty space before the newline and after it.