Home > Java > javaTutorial > What Does \'? extends HasWord\' Mean in Java Generics?

What Does \'? extends HasWord\' Mean in Java Generics?

Mary-Kate Olsen
Release: 2024-11-28 11:12:11
Original
436 people have browsed it

What Does

Understanding the Question Mark in Java Generics Type Parameters

In Java generics, you may encounter type parameters with question marks, like "? extends HasWord". This syntax indicates a bounded wildcard type. Let's delve into its meaning:

"? extends HasWord" means that the parameter represents a class or interface that extends HasWord. It encompasses HasWord itself as well as all its subclasses. In other words, it allows for any type that satisfies the condition "instanceof HasWord" or accepts null.

This concept originates from Item 31 of Effective Java 3rd Edition, "Use bounded wildcards to increase flexibility and safety". Essentially, a bounded wildcard widens the acceptable types beyond what a non-parameterized type could accept.

For example, if a method signature requires a List, passing in a List would not be allowed. However, if the signature used List, the ChildOfHasWord list would be acceptable.

The subtle difference between "? extends HasWord" and "? super HasWord" is also worth noting. The former is suitable when you need a collection from which data will be retrieved, while the latter is appropriate for collections where data will be added. This guideline is often referred to as "PEGS": Producer-extends, Consumer-super.

In summary, the question mark in Java generics type parameters, "? extends HasWord", indicates a bounded wildcard type that represents any class or interface extending HasWord, offering greater flexibility and improved type safety. Understanding this concept enhances your grasp of Java generics and enables you to optimize code design for efficiency and maintainability.

The above is the detailed content of What Does \'? extends HasWord\' Mean in Java Generics?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template