Illegal escape character error in Java regex

Question

I've read the manual, and at the end there was an exercise:

Use a backreference to write an expression that will match a person's name only if that person's first name and last name are the same.

I've written the next program http://pastebin.com/YkuUuP5M
But when I compile it, I'm getting an error:

PersonName.java:18: illegal escape character
p = Pattern.compile("([A-Z][a-zA-Z]+)\s+\1");
                                      ^

If I rewrite 18 line in this way:

pattern = Pattern.compile(console.readLine("%nEnter your regex: "));

and write the pattern in the console, then the program works fine. Why I can't use the pattern as in the 1st program case and is there some way to fix it?

Answer 1

You want to get this text into a string:

([A-Z][a-zA-Z]+)\s+\1

However, \ in a string literal in Java source code is the character used for escaping (e.g. "\t" for tab). Therefore you need to use "\" in a string literal to end up with a single backslash in the resulting string. So you want:

"([A-Z][a-zA-Z]+)\\s+\\1"

Note that there's nothing regular-expression-specific to this. Any time you want to express a string containing a backslash in a Java string literal, you'll need to escape that backslash. Regular expressions and Windows filenames are just the most common cases for that.