Java regular expression value.split(“\\.”), “the back slash dot” divides by character?

Question

From what I understand, the backslash dot (\\.) means one character of any character? So because backslash is an escape, it should be backslash backslash dot (

Answer 1

My guess is that you are missing that backslash ('\') characters are escape characters in Java String literals. So when you want to use a '\' escape in a regex written as a Java String you need to escape it; e.g.

Pattern.compile("\.");   // Java syntax error

// A regex that matches a (any) character
Pattern.compile(".");  

// A regex that matches a literal '.' character
Pattern.compile("\\.");  

// A regex that matches a literal '\' followed by one character
Pattern.compile("\\\\.");

The String.split(String separatorRegex) method splits a String into substrings separated by substrings matching the regex. So str.split("\\.") will split str into substrings separated by a single literal '.' character.

Answer 2

The regex "." would match any character as you state. However an escaped dot "\." would match literal dot characters. Thus 192.168.1.1 split on "\." would result in {"192", "168", "1", "1"}.

Your wording isn't completely clear, but I think this is what you're asking.