Seamus  Quitzon

Seamus Quitzon

1602637434

Integer Overflow in Java

While coding we may face some **integer overflow **cases for large inputs. Today I am going to discuss about two such scenarios which I have faced during problem solving.

Binary Search

Say you are given a sorted integer array and you are asked to find a particular element in that array. Since the array is sorted we can apply binary search algorithm to find the target element in O(logN) time; here N is the size of the input array.

The basic idea of binary search is each time we divide the array in half, find the mid index and compare the target element with the middle element. If the target element is greater than the middle element in that case we need to search only the right part of the array starting from the next index of the mid index. If the target element is less than the middle element then we need to search the left part of the array which ends before the mid index.

So the question is how we are going to find the mid index. It’s very simple; we need to maintain two pointers: _start _and _end; start _pointer is initialized with 0 and end pointer is initialized with N-1(N is the size of the array). So the formula we are going to use to find the mid index is:

mid = (start + end)/2.

At first glance this formula might seem innocent, but if we closely look at this formula we can see at first part it adds two integer numbers. Since we are adding two integer numbers there is a possibility of integer overflow if these two numbers are very large. So we should avoid this formula. Instead we can use the below formula to calculate the mid index:

mid = start + (end-start)/2;

Clearly “(end-start)/2” is less than “end” and this formula will not cause integer overflow for large values of start and end.

#problem-solving #sorting #integer-overflow #java #binary-search

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Integer Overflow in Java
Tyrique  Littel

Tyrique Littel

1600135200

How to Install OpenJDK 11 on CentOS 8

What is OpenJDK?

OpenJDk or Open Java Development Kit is a free, open-source framework of the Java Platform, Standard Edition (or Java SE). It contains the virtual machine, the Java Class Library, and the Java compiler. The difference between the Oracle OpenJDK and Oracle JDK is that OpenJDK is a source code reference point for the open-source model. Simultaneously, the Oracle JDK is a continuation or advanced model of the OpenJDK, which is not open source and requires a license to use.

In this article, we will be installing OpenJDK on Centos 8.

#tutorials #alternatives #centos #centos 8 #configuration #dnf #frameworks #java #java development kit #java ee #java environment variables #java framework #java jdk #java jre #java platform #java sdk #java se #jdk #jre #open java development kit #open source #openjdk #openjdk 11 #openjdk 8 #openjdk runtime environment

Samanta  Moore

Samanta Moore

1620458875

Going Beyond Java 8: Local Variable Type Inference (var) - DZone Java

According to some surveys, such as JetBrains’s great survey, Java 8 is currently the most used version of Java, despite being a 2014 release.

What you are reading is one in a series of articles titled ‘Going beyond Java 8,’ inspired by the contents of my book, Java for Aliens. These articles will guide you step-by-step through the most important features introduced to the language, starting from version 9. The aim is to make you aware of how important it is to move forward from Java 8, explaining the enormous advantages that the latest versions of the language offer.

In this article, we will talk about the most important new feature introduced with Java 10. Officially called local variable type inference, this feature is better known as the **introduction of the word **var. Despite the complicated name, it is actually quite a simple feature to use. However, some observations need to be made before we can see the impact that the introduction of the word var has on other pre-existing characteristics.

#java #java 11 #java 10 #java 12 #var #java 14 #java 13 #java 15 #verbosity

Samanta  Moore

Samanta Moore

1620462686

Spring Boot and Java 16 Records

In this article, we will discuss Java 16’s newest feature, Records. Then we will apply this knowledge and use it in conjunction with a Spring Boot application.

On March 16th, 2021, Java 16 was GA. With this new release, tons of new exciting features have been added. Check out the release notes to know more about these changes in detail. This article’s focus will be on Java Records, which got delivered with JEP 395. Records were first introduced in JDK 14 as a preview feature proposed by JEP 359, and with JDK 15, they remained in preview with JEP 384. However, with JDK 16, Records are no longer in preview.

I have picked Records because they are definitely the most favored feature added in Java 16, according to this Twitter poll by Java Champion Mala Gupta.

I also conducted a similar survey, but it was focused on features from Java 8 onwards. The results were not unexpected, as Java 8 is still widely used. Very unfortunate, though, as tons of new features and improvements are added to newer Java versions. But in terms of features, Java 8 was definitely a game-changer from a developer perspective.

So let’s discuss what the fuss is about Java Records.

#java #springboot #java programming #records #java tutorials #java programmer #java records #java 16

Seamus  Quitzon

Seamus Quitzon

1602637135

Learning by Doing: How to Learn Java Basics by Building Your Own Project

Java is not the hardest language to start with. So, it becomes way popular among novice developers joining the ranks of Java coders every single day. If you are reading this blog post, you might be interested in learning Java.

Java is widely used across industry, and especially in the area of Enterprise software, which results in many high paying job opportunities and makes this programming language a common language for newbies. A general promotion of it within colleges and other institutions providing a formal Computer Science education also contributes to its popularity.

However, these are not the only advantages of Java — among other things, it allows you to adopt good practices and makes it way easier to learn other languages in the future. And with no doubt, you can easily learn it if you’re following the right approach. In this post, I am going to share some of them with you.

The Importance of Practice in Programming

Beyond all doubt, practice is important and valuable. But, before we get to the advantages of hands-on experience, I want to draw your attention to one essential thing I often tell my students.

New programmers who are just learning and start implementing things, without being supervised, often end up adapting bad practices. To avoid that, especially when you are making your first steps in programming, I recommend looking for a person who will supervise you and teach you. A strong mentorship with someone engaged in a serious project, as well as communication within the community in the form of sharing code and asking for feedback, is worth the effort. Similarly, when you are applying for your first job, you want to be looking for a company with a strong team and a good leader who would be keen on investing into your learning.

Now, let’s return to practical experience. Learning by doing is different from learning by passively consuming the information. To make sure we can use all the newly acquired technology, we should put our skills to test and write tons of code. The benefits of hands-on experience are almost endless.

Efficiency and Productivity

By practicing, you get a clear understanding of what programming is. Consequently, you start doing better with each new hands-on task, complete it faster, and thus become more productive.

Even if you are not working on real-world projects yet, it’s important to get used to having deadlines. They are inextricably linked to the programming process. My recommendation is to set up your own deadlines while practicing stage and follow them as closely as possible.

#java #learn java #java code #learn java in easy way #learn java course #learn java development

Seamus  Quitzon

Seamus Quitzon

1599278213

Java Integer Cache: Why Integer.valueOf(127) == Integer.valueOf(127) Is True

Short Answer

The short answer to this question is, direct assignment of an int literal to an Integer reference is an example of auto-boxing concept where the literal value to object conversion code is handled by the compiler, so during compilation phase compiler converts Integer a = 127; to Integer a = Integer.valueOf(127);.

The Integer class maintains an internal IntegerCache for integers which, by default, ranges from -128 to 127 and Integer.valueOf() method returns objects of mentioned range from that cache. So a == b returns true because a and b both are pointing to the same object.

Long Answer

In order to understand the short answer, let’s first understand the Java types, all types in Java lies under two categories

  1. Primitive Types: There are 8 primitive types (byteshortintlongfloatdoublechar, and boolean) in Java, which holds their values directly in the form of binary bits.
  2. For example, int a = 5; int b = 5; here a and b directly holds the binary value of 5, and if we try to compare a and b using a == b, we are actually comparing 5 == 5, which returns true.
  3. Reference Types: All types other than primitive types lies under the category of reference types, e.g. Classes, Interfaces, Enums, Arrays, etc. and reference types holds the address of the object instead of the object itself.
  4. For example, Integer a = new Integer(5); Integer b = new Integer(5), here, a and b do not hold the binary value of 5 instead a and b holds memory addresses of two separate objects where both objects contain a value 5. So if we try to compare a and b using a == b, ,we are actually comparing those two separate memory addresses. Hence, we get false, to perform actual equality on a and b we need to perform a.euqals(b).

Reference types are further divided into 4 categories:  Strong, Soft, Weak and Phantom References.

And we know that Java provides wrapper classes for all primitive types and support auto-boxing and auto-unboxing.

// Example of auto-boxing, here c is a reference type

Integer c = 128; // Compiler converts this line to Integer c = Integer.valueOf(128);

// Example of auto-unboxing, here e is a primitive type

int e = c; // Compiler converts this line to int e = c.intValue();


Now, if we create two integer objects `a` and `b,` and try to compare them using the equality operator `==`, we will get `false` because both references are holding different-different objects

Integer a = 128; // Compiler converts this line to Integer a = Integer.valueOf(128);

Integer b = 128; // Compiler converts this line to Integer b = Integer.valueOf(128);

System.out.println(a == b); // Output – false


But if we assign the value `127` to both `a` and `b` and try to compare them using the equality operator `==`, we will get `true` why?

Integer a = 127; // Compiler converts this line to Integer a = Integer.valueOf(127);

Integer b = 127; // Compiler converts this line to Integer b = Integer.valueOf(127);

System.out.println(a == b); // Output – true


As we can see in the code, we are assigning different objects to `a` and `b` but `a == b` can return true only if both `a` and `b` are pointing to the same object.

So, how does the comparison return true? what's actually happening here? are `a` and `b` pointing to the same object?

Well, until now, we know that the code `Integer a = 127;` is an example of auto-boxing and compiler automatically converts this line to `Integer a = Integer.valueOf(127);`.

So, it is the `Integer.valueOf()` method that is returning these integer objects, which means this method must be doing something under the hood.

And if we take a look at the source code of `Integer.valueOf()` method, we can clearly see that if the passed int literal `i` is greater than `IntegerCache.low` and less than `IntegerCache.high ,`then the method returns Integer objects from `IntegerCache`. Default values for `IntegerCache.low` and `IntegerCache.high` are `-128` and `127` respectively.

In other words, instead of creating and returning new integer objects, `Integer.valueOf()` method returns Integer objects from an internal `IntegerCache` if the passed `int` literal is greater than `-128` and less than `127`.

#java #cache #short #long #btye #integer cache #java integer