Essential Java Tips and Tricks for Programmers


Java is a powerful and versatile programming language. It is used in various applications. It's known for its reliability, portability, and safety, which makes it a famous desire for developers. Java is likewise exceptionally easy to analyze, which makes it a terrific desire for novices.

However, it's critical to remember that simplicity might operate as an obstacle. If you're not mindful, you could be caught by Java's accessibility and neglect to explore the unique opportunities the language offers.

Let's have a look at some tips to assist you develop as a Java developer and improve your language proficiency.

Tip 1: Never Ignore The Basics

To be a great Java developer, you must be ace at knowing the fundamentals first. Many concepts such as "syntax," "data types," "operators," "control flow," and "object-oriented programming concepts" should be on your learning agenda. Where to learn? You have a number of resources available in the market be it- as books, online courses, and tutorials. All of these resources will help you understand the fundamentals.

You have to go beyond just knowing the syntax and also enhance your problem-solving abilities. This way, you might advance as a developer.

Tip 2: Know The Source Code and Algorithm

It is critical to comprehend the source code and algorithm of any program you are working on. Simple programs with only one "if else" statement will demonstrate this. You will better comprehend how the program functions and solve any issues that arise if you are familiar with the source code.

It can be beneficial to lay out the source code on paper if you are a beginner. That is before beginning to code. This could make it easier as a way to see how the program flows and spot any possible issues.

You can start breaking the program into smaller components. That is when you've mastered the source code. This can help you understand how the program operates and solve any potential issues more easily.

Tip 3: Always Remember to Allocate Memory

In Java, memory is allocated at runtime.

You must employ the new keyword when you are allocating memory for objects, You will receive a NullPointerException if the new keyword is not employed. The motive for this is that no memory will have been set aside for the object because it hasn't yet been created.

Let us take an example to better understand this. The below code will not work:

int[] array = {1, 2, 3};

There are no compilation issues with this example. At runtime, it will nevertheless raise a NullPointerException. No memory is set aside for the array. It is because the memory hasn't been created yet.

To fix this code, you need to employ the new keyword:

int[] array = new int[3];

This code will allocate memory for the array at runtime, and the NullPointerException will be avoided.

Tip 4: Refrain From Creating Unneeded Objects

When you create an object in Java, you are reserving space in the memory of the computer for that object. The space gets employed to store the object's data and methods.

The more objects you create, the more memory you will employ. This will lead to slow ing down of your program. It can even lead to out-of-memory errors.

To avoid memory leaks, you should be mindful of the number of objects you create in your Java code.

Tip 5: Primitive Classes Should Be Chosen Above Wrapper Classes

Objects that represent primitive data types are called wrapper classes. They are frequently used to provide primitive data types further features, including the capacity to store null values. Wrapper classes need more memory and processing power than primitive data types, hence they may be slower than those latter.

Example

The given Java code defines a class with a main method that compares two sets of integers: the primitive variables num_1 and num_2, which have the same value of 10, resulting in the first print statement displaying "true."

Similarly, the Integer wrapper objects wrapnum_1 and wrapnum_2 are auto-boxed with the same value, leading to the second print statement also displaying "true" as their values are compared using the equals method.

Algorithm

  • Step 1: Declare two int variables, num_1 and num_2. Then, initialize them to 10.

  • Step 2: Declare two Integer variables, wrapnum_1 and wrapnum_2. Then, initialize them to 10.

  • Step 3: Employ the == operator to compare num_1 and num_2. This will compare the reference values of the two variables. Since the two variables reference the same object, the output will be true.

  • Step 4:Employ the equals() method to compare wrapnum_1 and wrapnum_2. Note: It will compare the values of the two variables. Because the two variables have the same value, the output will be true.

Example

public class WrapperComparison {
    public static void main(String[] args) {
        int num_1 = 10;
        int num_2 = 10;
        Integer wrapnum_1 = 10;
        Integer wrapnum_2 = 10;

        System.out.println(num_1 == num_2);      // Output: true
        System.out.println(wrapnum_1.equals(wrapnum_2)); // Output: true
    }
}

Output

true
true

Tip 6: Strings Management

Strings are objects that cannot be altered in Java. Therefore, once you produce them, they cannot be altered. Java creates a new string object whenever you concatenate two strings. The new string object contains the concatenation of the two original strings. It can lead to significant memory overhead. (Particularly if you are concatenating strings frequently.)

How do you prevent this? You instantly create a string object, by employing the String literal syntax.

For example, the below code will create a string object without creating a new object:

Syntax

String fast = "This string is better"

for()

The code first creates two strings, slow and fast. Then, it starts a timer and appends the number 0 to the end of slow 10000 times. It then stops the timer and prints the time taken to execute this operation. Finally, it repeats the same operation for fast and prints the time taken.

Algorithm

  • Step 1: Create two strings, slow and fast.

  • Step 2: Initialize a variable start to the current time in milliseconds.

  • Step 3: Iterate 10000 times: Concatenate the string slow with the string " " and the value of i.

  • Step 4: Set the variable slowEndTime to the current time in milliseconds.

  • Step 5: Iterate 10000 times: Concatenate the string fast with the string " " and the value of i.

  • Step 6: Set the variable fastEndTime to the current time in milliseconds.

  • Step 7: Print the time taken to execute the slow string.

  • Step 8: Print the time taken to execute the fast string.

Example

public class StringPerformance {

    public static void main(String[] args) {
        String slow = "This is a slow string.";
        String fast = "This is a fast string.";

        long start = System.currentTimeMillis();

        for (int i = 0; i < 10000; i++) {
            slow += " " + i;
        }

        long slowEndTime = System.currentTimeMillis();

        for (int i = 0; i < 10000; i++) {
            fast += " " + i;
        }

        long fastEndTime = System.currentTimeMillis();

        System.out.println("Time taken to execute slow string: " + (slowEndTime - start) + " milliseconds");
        System.out.println("Time taken to execute fast string: " + (fastEndTime - slowEndTime) + " milliseconds");
    }
}

Output

Time taken to execute slow string: 140 milliseconds
Time taken to execute fast string: 23 milliseconds

Tip 7: Keep Coding

Java is constantly evolving. Java proficiency requires a love for learning and an ongoing commitment to improvement. The time and effort required to acquire the language must also be willingly expended. As you practice more, you'll improve.

Conclusion

For developing into a skilled Java programmer, remember the advice from above. They will help you develop your Java proficiency and skills, which will increase your employability.

Updated on: 29-Aug-2023

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