[regex] What is a word boundary in regex, does \b match hyphen '-'?

I'm trying to use regexes to match space-separated numbers. I can't find a precise definition of \b ("word boundary"). I had assumed that -12 would be an "integer word" (matched by \b\-?\d+\b) but it appears that this does not work. I'd be grateful to know of ways of .

[I am using Java regexes in Java 1.6]

Example:

Pattern pattern = Pattern.compile("\\s*\\b\\-?\\d+\\s*");
String plus = " 12 ";
System.out.println(""+pattern.matcher(plus).matches());

String minus = " -12 ";
System.out.println(""+pattern.matcher(minus).matches());

pattern = Pattern.compile("\\s*\\-?\\d+\\s*");
System.out.println(""+pattern.matcher(minus).matches());

This returns:

true
false
true

A word boundary is a position that is either preceded by a word character and not followed by one, or followed by a word character and not preceded by one.

I would like to explain Alan Moore's answer

A word boundary is a position that is either preceded by a word character and not followed by one or followed by a word character and not preceded by one.

Suppose I have a string "This is a cat, and she's awesome", and I am supposed to replace all occurrence(s) the letter 'a' only if this letter exists at the "Boundary of a word", i.e. the letter a inside 'cat' should not be replaced.

So I'll perform regex (in Python) as

re.sub("\ba","e", myString.strip()) //replace a with e

so the output will be This is e cat end she's ewesome

when you use \\b(\\w+)+\\b that means exact match with a word containing only word characters ([a-zA-Z0-9])

in your case for example setting \\b at the begining of regex will accept -12(with space) but again it won't accept -12(without space)

for reference to support my words: https://docs.oracle.com/javase/tutorial/essential/regex/bounds.html

I ran into an even worse problem when searching text for words like .NET, C++, C#, and C. You would think that computer programmers would know better than to name a language something that is hard to write regular expressions for.

Anyway, this is what I found out (summarized mostly from http://www.regular-expressions.info, which is a great site): In most flavors of regex, characters that are matched by the short-hand character class \w are the characters that are treated as word characters by word boundaries. Java is an exception. Java supports Unicode for \b but not for \w. (I'm sure there was a good reason for it at the time).

The \w stands for "word character". It always matches the ASCII characters [A-Za-z0-9_]. Notice the inclusion of the underscore and digits (but not dash!). In most flavors that support Unicode, \w includes many characters from other scripts. There is a lot of inconsistency about which characters are actually included. Letters and digits from alphabetic scripts and ideographs are generally included. Connector punctuation other than the underscore and numeric symbols that aren't digits may or may not be included. XML Schema and XPath even include all symbols in \w. But Java, JavaScript, and PCRE match only ASCII characters with \w.

Which is why Java-based regex searches for C++, C# or .NET (even when you remember to escape the period and pluses) are screwed by the \b.

Note: I'm not sure what to do about mistakes in text, like when someone doesn't put a space after a period at the end of a sentence. I allowed for it, but I'm not sure that it's necessarily the right thing to do.

Anyway, in Java, if you're searching text for the those weird-named languages, you need to replace the \b with before and after whitespace and punctuation designators. For example:

public static String grep(String regexp, String multiLineStringToSearch) {
    String result = "";
    String[] lines = multiLineStringToSearch.split("\\n");
    Pattern pattern = Pattern.compile(regexp);
    for (String line : lines) {
        Matcher matcher = pattern.matcher(line);
        if (matcher.find()) {
            result = result + "\n" + line;
        }
    }
    return result.trim();
}

Then in your test or main function:

    String beforeWord = "(\\s|\\.|\\,|\\!|\\?|\\(|\\)|\\'|\\\"|^)";   
    String afterWord =  "(\\s|\\.|\\,|\\!|\\?|\\(|\\)|\\'|\\\"|$)";
    text = "Programming in C, (C++) C#, Java, and .NET.";
    System.out.println("text="+text);
    // Here is where Java word boundaries do not work correctly on "cutesy" computer language names.  
    System.out.println("Bad word boundary can't find because of Java: grep with word boundary for .NET="+ grep("\\b\\.NET\\b", text));
    System.out.println("Should find: grep exactly for .NET="+ grep(beforeWord+"\\.NET"+afterWord, text));
    System.out.println("Bad word boundary can't find because of Java: grep with word boundary for C#="+ grep("\\bC#\\b", text));
    System.out.println("Should find: grep exactly for C#="+ grep("C#"+afterWord, text));
    System.out.println("Bad word boundary can't find because of Java:grep with word boundary for C++="+ grep("\\bC\\+\\+\\b", text));
    System.out.println("Should find: grep exactly for C++="+ grep(beforeWord+"C\\+\\+"+afterWord, text));

    System.out.println("Should find: grep with word boundary for Java="+ grep("\\bJava\\b", text));
    System.out.println("Should find: grep for case-insensitive java="+ grep("?i)\\bjava\\b", text));
    System.out.println("Should find: grep with word boundary for C="+ grep("\\bC\\b", text));  // Works Ok for this example, but see below
    // Because of the stupid too-short cutsey name, searches find stuff it shouldn't.
    text = "Worked on C&O (Chesapeake and Ohio) Canal when I was younger; more recently developed in Lisp.";
    System.out.println("text="+text);
    System.out.println("Bad word boundary because of C name: grep with word boundary for C="+ grep("\\bC\\b", text));
    System.out.println("Should be blank: grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));
    // Make sure the first and last cases work OK.

    text = "C is a language that should have been named differently.";
    System.out.println("text="+text);
    System.out.println("grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));

    text = "One language that should have been named differently is C";
    System.out.println("text="+text);
    System.out.println("grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));

    //Make sure we don't get false positives
    text = "The letter 'c' can be hard as in Cat, or soft as in Cindy. Computer languages should not require disambiguation (e.g. Ruby, Python vs. Fortran, Hadoop)";
    System.out.println("text="+text);
    System.out.println("Should be blank: grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));

P.S. My thanks to http://regexpal.com/ without whom the regex world would be very miserable!

I talk about what \b-style regex boundaries actually are here.

The short story is that they’re conditional. Their behavior depends on what they’re next to.

# same as using a \b before:
(?(?=\w) (?<!\w)  | (?<!\W) )

# same as using a \b after:
(?(?<=\w) (?!\w)  | (?!\W)  )

Sometimes that isn’t what you want. See my other answer for elaboration.

I believe that your problem is due to the fact that - is not a word character. Thus, the word boundary will match after the -, and so will not capture it. Word boundaries match before the first and after the last word characters in a string, as well as any place where before it is a word character or non-word character, and after it is the opposite. Also note that word boundary is a zero-width match.

One possible alternative is

(?:(?:^|\s)-?)\d+\b

This will match any numbers starting with a space character and an optional dash, and ending at a word boundary. It will also match a number starting at the beginning of the string.

In the course of learning regular expression, I was really stuck in the metacharacter which is \b. I indeed didn't comprehend its meaning while I was asking myself "what it is, what it is" repetitively. After some attempts by using the website, I watch out the pink vertical dashes at the every beginning of words and at the end of words. I got it its meaning well at that time. It's now exactly word(\w)-boundary.

My view is merely to immensely understanding-oriented. Logic behind of it should be examined from another answers.

A word boundary can occur in one of three positions:

1. Before the first character in the string, if the first character is a word character.
2. After the last character in the string, if the last character is a word character.
3. Between two characters in the string, where one is a word character and the other is not a word character.

Word characters are alpha-numeric; a minus sign is not. Taken from Regex Tutorial.

Word boundary \b is used where one word should be a word character and another one a non-word character. Regular Expression for negative number should be

--?\b\d+\b

check working DEMO

Check out the documentation on boundary conditions:

http://java.sun.com/docs/books/tutorial/essential/regex/bounds.html

Check out this sample:

public static void main(final String[] args)
    {
        String x = "I found the value -12 in my string.";
        System.err.println(Arrays.toString(x.split("\\b-?\\d+\\b")));
    }

When you print it out, notice that the output is this:

[I found the value -, in my string.]

This means that the "-" character is not being picked up as being on the boundary of a word because it's not considered a word character. Looks like @brianary kinda beat me to the punch, so he gets an up-vote.

I think it's the boundary (i.e. character following) of the last match or the beginning or end of the string.

Reference: Mastering Regular Expressions (Jeffrey E.F. Friedl) - O'Reilly

\b is equivalent to (?<!\w)(?=\w)|(?<=\w)(?!\w)