Saturday, 19 August 2017

New Regex Features in Java 9

In this article I will focus on the regular expression features that are specific to Java 9 and were not available in earlier version of the JDK. There is not many, though.

Java 9 Regular Expression Module

The JDK in Java 9 is split up into modules. One could rightfully expect that there is a new module for the regular expression handling packages and classes. Actually there is none. The module java.base is the default module on which all other modules depend on by default and thus the classes of the exported packages are always available in Java applications. The regular expression package java.util.regex is exported by this module. This makes the development a bit simpler: there is no need to explicitly ‘require’ a module if we want to use regular expressions in our code. It seems that regular expressions are so essential to Java that it got included in the base module.

Thursday, 17 August 2017

Java 8 Optionals

While programming, we have all faced the most (in)famous NullPointerException. And I believe we all would agree that encountering NullPointerException is also a pain. Just to keep the readers informed, the famous computer scientist Tony Hoare introduced null references and he thinks this as A million-dollar mistake. We all know, it’s very easy to implement but it’s quite unpredictable as well. And that’s why developers need to be very cautious.

Wednesday, 16 August 2017

Java 8 Default Methods On Interface : What, Why and Example

Whenever someone talks about Java 8, the first thing he speaks about is lambda expression and How lambda expression will change the way you use Collections API today. In truth, lambda expression would not be that useful had language not been enhanced to support default methods on Java Interface. Also known as a virtual extension or defender methods, they allow you to declare a non-abstract method inside Java interface. Which means, finally you can add new methods without breaking all classes, which implements a certain interface. This opens a new path for enhancing and evolving existing Collection API to take advantage of lambda expressions. For example, now you can iterate over all elements of Collection in just one line, as opposed to four lines it requires you to do prior to Java 8.

Friday, 11 August 2017

How to create a thread-safe ConcurrentHashSet in Java 8?

Until JDK 8, there was no way to create a large, thread-safe, ConcurrentHashSet in Java. The java.util.concurrent package doesn't even have a class called ConcurrentHashSet, but from JDK 8 onwards, you can use newly added keySet(default value) and newKeySet() method to create a ConcurrentHashSet backed by ConcurrentHashMap. Unlike tactical solutions like using concurrent hash map with dummy value or using the set view of the map, where you cannot add new elements. The Set returned by keySet(defaultValue) and newKeySet() method of JDK 8 is a proper set, where you can also add new elements along with performing other set operations e.g. contains(), remove() etc. Though you need to be careful that these methods are only available in ConcurrentHashMap class and not in ConcurrentMap interface, so you need to use a ConcurrentHashMap variable to hold the reference, or you need to use type casting to cast a ConcurrentHashMap object stored in ConcurrentMAp variable.

Thursday, 10 August 2017

How to determine if a string has all unique characters in Java?

This is one of the common string based coding problem from programming job interviews. You need to write a program to determine if a given string has all unique characters or not. For example input= "Java" then your function should return false because all characters are not unique, and if the input is "Python" then your program should return true because all characters in Python are unique. For the purpose of this problem, you can assume the given String only contains ASCII printable characters, though you should always verify that with the interviewer. You can also assume that your solution needs to case-sensitive i.e. "P" and "p" will be considered two different characters and a string containing a letter in both capital and the small case will be considered unique. You are also free to solve the problem in place of using any additional data structure.