web.xml中配置的class老师是如何获得的呢？

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
	http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
  <display-name>Spring MVC Study</display-name>
  
    <!-- Spring应用上下文， 理解层次化的ApplicationContext -->
  <context-param>
 		<param-name>contextConfigLocation</param-name>
		<param-value>/WEB-INF/configs/spring/applicationContext*.xml</param-value>
  </context-param>
  
  <listener>
		<listener-class>
			org.springframework.web.context.ContextLoaderListener
		</listener-class>
  </listener>
  
    <!-- DispatcherServlet, Spring MVC的核心 -->
<servlet>
    <servlet-name>mvc-dispatcher</servlet-name>
    <servlet-class> org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <!-- DispatcherServlet对应的上下文配置， 默认为/WEB-INF/$servlet-name$-servlet.xml -->
		<init-param>
		    <param-name>contextConfigLocation</param-name>
		    <param-value>/WEB-INF/configs/spring/mvc-dispatcher-servlet.xml</param-value>
		</init-param>
		<load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>mvc-dispatcher</servlet-name>
    <!-- mvc-dispatcher拦截所有的请求-->
    <url-pattern>/</url-pattern>
</servlet-mapping>
  
</web-app>

上面的listener-class和servlet-class的值，我们在刚配置时时通过什么途径获取的呢？