java, regular expression, need to escape backslash in regex

Question

With reference to below question - String.replaceAll single backslashes with double backslashes

I wrote a test program, and I found that the result is true in both cases, whether I escape the backslash or not. This may be because - \t is a recognized Java String escape sequence. (Try \s and it would complain). - \t is taken as literal tab in the regex. I am somewhat unsure of the reasons.

Is there any general guideline about escaping regex in Java. I think using two backslashes is the correct approach.

I would still like to know your opinions.

public class TestDeleteMe {

  public static void main(String args[]) {
    System.out.println(System.currentTimeMillis());

    String str1 = "a    b"; //tab between a and b 

    //pattern - a and b with any number of spaces or tabs between 
    System.out.println("matches = " + str1.matches("^a[ \\t]*b$")); 
    System.out.println("matches = " + str1.matches("^a[ \t]*b$")); 
  }
}

Answer 1

There are two interpretations of escape sequences going on: first by the Java compiler, and then by the regexp engine. When Java compiler sees two slashes, it replaces them with a single slash. When there is t following a slash, Java replaces it with a tab; when there is a t following a double-slash, Java leaves it alone. However, because two slashes have been replaced by a single slash, regexp engine sees \t, and interprets it as a tab.

I think that it is cleaner to let the regexp interpret \t as a tab (i.e. write "\\t" in Java) because it lets you see the expression in its intended form during debugging, logging, etc. If you convert Pattern with \t to string, you will see a tab character in the middle of your regular expression, and may confuse it for other whitespace. Patterns with \\t do not have this problem: they will show you a \t with a single slash, telling you exactly the kind of whitespace that they match.