Java Regular Expression value.split("\\."), The Forward Slash Dot divides by character?

Question

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

What does this do to a string. I just saw this in an existing code I am working on. From what I understand it will split the string into individual characters. Why do this instead of String.toCharArray (I just guessed this name I know there is a Java string function to split string to char array). So this splits the String to an array of string which contains only one char for each string in the array?

Any Java Gurus out there that used this before?

Answer 1

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.

Answer 2

Your question is badly worded, but 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) one 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 literal '.' character.