Java Top

I just announced the new Learn Spring course, focused on the fundamentals of Spring 5 and Spring Boot 2:


1. Overview

In this short tutorial, we'll discuss different ways to compare two Long instances. We emphasize the problems that arise when using the reference comparison operator (==).

2. Problem Using Reference Comparison

Long is a wrapper class for the primitive type long. Since they are objects and not primitive values, we need to compare the content of Long instances using .equals() instead of the reference comparison operator (==).

In some cases, we may get the idea that == is okay, but looks are deceiving. Consider that we can use == with low numbers:

Long l1 = 127L;
Long l2 = 127L;

assertThat(l1 == l2).isTrue();

but not with larger numbers. We would end up having issues if values are out of the range -128 to 127, having a completely different and unexpected result:

Long l1 = 128L;
Long l2 = 128L;

assertThat(l1 == l2).isFalse();

This is because Java maintains a constant pool for instances of Long between -128 and 127.

This optimization, though, does not give us a license to use ==. In the general case, two boxed instances having the same primitive value don't yield the same object reference.

3. Using .equals()

One of the solutions is to use the .equals(). This will evaluate the content (and not the reference) of both objects:

Long l1 = 128L;
Long l2 = 128L;


4. Objects.equals()

The problem with using equals() is that we need to be cautious not to call it on the null reference.

Luckily, there's a null-safe utility method we can use – Objects.equals().

Let's see how it works in practice:

Long l1 = null;
Long l2 = 128L;

assertThatCode(() -> Objects.equals(l1, l2)).doesNotThrowAnyException();

As we can see, we don't need to bother if any of the Longs we want to compare is null.

Under the hood, Objects.equals() first uses the == operator for the comparison, and if that fails it uses a standard equals().

5. Unboxing Long Values

5.1. Using the .longValue() Method

Next, let's use the “==” comparison operator, but in a safe way. The class Number has a method .longValue() which unwraps the primitive long value:

Long l1 = 128L;
Long l2 = 128L;

assertThat(l1.longValue() == l2.longValue()).isTrue();

5.2. Casting to Primitive Values

A different way to unbox a Long is by casting the objects to primitive types. Therefore, we'll extract the primitive value and then we can proceed to use the comparison operator:

Long l1 = 128L;
Long l2 = 128L;

assertThat((long) l1 == (long) l2).isTrue();

Note that, for the .longValue() method or using casting, we should check if the object is null. We could have a NullPointerException if the object is null.

6. Conclusion

In this short tutorial, we have explored different options on how to compare Long objects. We have analyzed the differences when comparing references to objects or content.

As always, the full source code of the article is available over on GitHub.

Java bottom

I just announced the new Learn Spring course, focused on the fundamentals of Spring 5 and Spring Boot 2:

Inline Feedbacks
View all comments
7 months ago

How about

java.util.Objects.equals(l1, l2);


Loredana Crusoveanu
7 months ago
Reply to  scrooge

Yeah, using Objects.equals() is a good idea.
Under the hood, it uses both approaches – reference comparison operator and a null safe equals().

Comments are closed on this article!