Mapping a specific servlet to be the default servlet in Tomcat

This should be useful to you.

From the Java™ Servlet Specification Version 3.1 (JSR 340)

Chapter 12. Mapping Requests to Servlets

12.2 Specification of Mappings

In the Web application deployment descriptor, the following syntax is used to define mappings:

• A string beginning with a / character and ending with a /* suffix is used for path mapping.

• A string beginning with a *. prefix is used as an extension mapping.

• The empty string ("") is a special URL pattern that exactly maps to the application's context root, i.e., requests of the form http://host:port/<contextroot>/. In this case the path info is / and the servlet path and context path is empty string ("").

• A string containing only the / character indicates the "default" servlet of the application. In this case the servlet path is the request URI minus the context path and the path info is null.

• All other strings are used for exact matches only.

As an addition, read this nice explanation with short examples from the book Head First Servlets & JSP: Passing the Sun Certified Web Component Developer Exam (2nd edition) (quote):

The THREE types of <url-pattern> elements

1) EXACT match

Example:
<url-pattern>/Beer/SelectBeer.do</url-pattern>

• MUST begin with a slash (/).
• Can have an extension (like .do), but it’s not required.

2) DIRECTORY match

Example:
<url-pattern>/Beer/*</url-pattern>

• MUST begin with a slash (/).
• Always ends with a slash/asterisk (/*).

3) EXTENSION match

Example:
<url-pattern>*.do</url-pattern>

• MUST begin with an asterisk (*) (NEVER with a slash).
• After the asterisk, it MUST have a dot extension (.do, .jsp, etc.).

IMPORTANT NOTE:
The URL patterns represent logical / virtual structure, i.e. the patterns (paths) specified does not need to exist physically.

UPDATE

If you want, as you say in your comment,

I want host:port to hit my servlet, not the default tomcat servlet

then see the solution here:
How do I make my web application be the Tomcat default application

In other words, what you want is a path without application context, which implies the application context of the Tomcat default application.

Quote from the above link:

In a standard Tomcat installation, you will notice that under the same directory (CATALINA_BASE)/webapps/, there is a directory called ROOT (the capitals are important, even under Windows). That is the residence of the current Tomcat default application, the one that is called right now when a user calls up http://myhost.company.com[:port]. The trick is to put your application in its place.

I am not sure did I understood what you want but probably intercept 404 is what you want to do, then redirect where you want.

I've came here to forum because I have strange problem with tomcat 7, mine is doing just what you want ;)

This is only one way how I can have root, EMPTY

<servlet-mapping>
    <servlet-name>Default</servlet-name>
    <url-pattern></url-pattern>
</servlet-mapping>

That way : anything is redirected to this servlet, including images, etc, for example, I open another page, this show this one, root, then I can see in log 4 more request to same page, 3 for css and one for image.

<servlet-mapping>
    <servlet-name>Default</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>