Font Awesome Free 5.13.0 by @fontawesome - License - (Icons: CC BY 4.0, Fonts: SIL OFL 1.1, Code: MIT License)
Skip to main content

Generics (Java)

An overview of generics in Java

Generics basics

Simple class hierarchy for examples:

public class Animal {}
public class Dog extends Animal {}

Example generic interface:

List<Dog> dogList = new ArrayList<Dog>();
dogList.add(new Dog());
Dog dog = dogList.get(0);
dogList.add(new Object()); // compiler error

Example generic method:

public <T extends Animal> T getFirstAnimal(List<T> animals) {
return animals.get(0);

T extends Animal is a type bound.


What if you need something representing any kind of list of animals?

List<Dog> dogList = new ArrayList<Dog>();
List<Animal> animalList = dogList; // compiler error

Reason why this fails: a proper List<Animal> allows adding any Animal, while a List<Dog> should only allow adding Dogs. This means that the two types are not compatible.

If we only care about the fact that our List contains some kind of Animals, we can use type wildcards to define this:

List<Dog> dogList = new ArrayList<Dog>();
List<? extends Animal> extendsAnimalList = dogList; // works

? extends Animal is called a subtype wildcard. ? super Dog is called a supertype wildcard

List<Dog> dogList = new ArrayList<Dog>();

List<? extends Animal> extendsAnimalList = dogList;
Animal animal = extendsAnimalList.get(0); // works
extendsAnimalList.add(new Dog()); // compiler error

List<? super Dog> superDogList = dogList;
Animal animal2 = superDogList.get(0); // compiler error
superDogList.add(new Dog()); // works

For very generic code, you can also use a wildcard (? without type bounds)

A note about arrays

We saw above that it is not possible to assign a List<Dog> to a List<Animal>, which makes sense. However, this is not the case for arrays. You can easily assign a Dog[] array to an Animal[] array without the compiler complaining. However, if you then attempt to insert an Animal that is not a Dog into the array, you will get an error at runtime.

Dog[] dogs = new Dog[10];
Animal[] animals = dogs; // works
animals[0] = new Animal() // fails at runtime

Generics inside the Java Virtual Machine

Type erasure

Java only added generics in version 1.5. Before that, instead of the generic ArrayList<T>, there was just the class ArrayList. When introducing generics, the Java team decided to maintain compatibility by actually erasing the generic type information at compile time, meaning that the byte code running in the Java Virtual Machine does not know anything about generics.

Example generic type:

public class AnimalWrapper<T extends Animal> {    
private T wrappedAnimal;

public AnimalWrapper(T wrappedAnimal) {
this.wrappedAnimal = wrappedAnimal;

public T getWrappedAnimal() {
return this.wrappedAnimal;

The above type is compiled into the following raw type:

public class AnimalWrapper {    
private Animal wrappedAnimal;

public AnimalWrapper(Animal wrappedAnimal) {
this.wrappedAnimal = wrappedAnimal;

public Animal getWrappedAnimal() {
return this.wrappedAnimal;

Before erasing the types, the compiler checks for errors involving generic types. For example, it will forbid wrapping an Animal in an AnimalWrapper<Dog>. This means that, although the types are erased later on, the type variables are still respected.

Cast insertion

Although the compiler checks for generic type mismatches, that this in itself is not always enough. An example is the following code:

List<Dog> dogList = new ArrayList<Dog>();        
List rawList = dogList;
Animal animal = new Animal();

The above code generates warnings, but if you choose to ignore those, you now have an Animal sitting inside a List<Dog>. This is known as heap pollution. But what about type safety?

Java handles this in the compiler by inserting a cast whenever the code reads from an expression with erased type. This means that, while we can add the Animal to our List<Dog>, we will get a ClassCastException if we try to retrieve that Animal as a Dog.

List<Dog> dogList = new ArrayList<Dog>();  
List rawList = dogList;
Animal animal = new Animal();

Animal retrievedAnimal = dogList.get(0); // works
Dog retrievedDog = dogList.get(0); // ClassCastException

Compiled code is equivalent to this:

List dogList = new ArrayList();  
List rawList = dogList;
Animal animal = new Animal();

// note that the erased List.get() method returns an Object
// cast insertion generates casts based on target type
Animal retrievedAnimal = (Animal) dogList.get(0); // works
Dog retrievedDog = (Dog) dogList.get(0); // ClassCastException

when we get the ClassCastException on the last line, that does not help us to find the actual source of the problem (which is the code where we inserted an Animal inside a List<Dog>). When debugging such problem, it can be useful to use a checked view of the List. This checks the type of inserted objects as they are inserted.

List<Dog> dogList = 
Collections.checkedList(new ArrayList<Dog>(), Dog.class);

List rawList = dogList;
Animal animal = new Animal();
rawList.add(animal); // ClassCastException

Note the use of Dog.class, a Class<Dog> instance which is needed to know the actual value of the type parameter for the List at runtime.

Bridge methods

In some cases, basic type erasure would lead to problems with method overriding. In order to prevent this, the Java compiler sometimes generates bridge methods.

Example class:

public class GoodBoyList extends ArrayList<Dog>{
public boolean add(Dog dog) {
return super.add(dog);

Now, let's say we use it this way:

GoodBoyList goodBoyList = new GoodBoyList();
ArrayList<Dog> dogList = goodBoyList;
dogList.add(new Dog());

After erasure, we have add(Dog) in GoodBoyList and add(Object) in ArrayList. The last line of the code above calls the add method on ArrayList. We expect the add method from GoodBoyList to override this, but the problem is the method signatures are different.

The compiler solves this by inserting a bridge method add(Object) into the GoodBoyList class. That method looks like this:

// overrides ArrayList.add(Object)
public boolean add(Object dog) {
return this.add((Dog) dog); // calls add(Dog)

Bridge methods can also be used when the return type varies. For example, imagine that our GoodBoyList also overrides the get(int) method.

public class GoodBoyList extends ArrayList<Dog>{
public Dog get(int i) {
Dog dog = super.get(i);
return dog;

Note: inside the Java Virtual Machine, a method is defined by its name, the number and types of its arguments and by its return type. This means that, after erasure, we again need a bridge method to make overriding work here. This way, we get two get methods in GoodBoyList:

  • Dog get(int): this is the actual method as defined in GoodBoyList
  • Object get(int): this is a generated bridge method that overrides the Object get(int) method in ArrayList.

The compiler takes care of the generation of bridge methods, so in principle you don’t have to worry about them. However, they may show up in stack traces or explain why the compiler complains about certain pieces of code.

The Class class

The Java language has a Class<T> class. A Class<T> object represent the class T. This class object can directly be obtained from the class T. It is also possible to to get a Class object from an instance of a class, but in that case you are getting the actual run-time type of that instance, which may be a subclass of its compile-time type.

Class<Dog> test = Dog.class; // ok
Class<Dog> test2 = new Dog().getClass(); // error
Class<? extends Dog> test3 = new Dog().getClass(); // ok

You can use the Class class to get more information regarding the value of a type variable at run-time (so after type erasure). Example:

public class Test<T> {    
Test(T object, Class<T> objectClass) {}

public class Main {
public static void main(String[] args) {
Dog dog = new Dog();
new Test<Animal>(dog, Animal.class);

T is erased, but we can still have a look at objectClass to check what kind of object we received.

The Class<T> object is also very useful when using reflection. For example, it can help you access the constructor(s) for the class.

Generics restrictions

Type arguments cannot be primitives

  • A type parameter must always be Object or a subclass of Object. This means that, for example, it is not possible to define an ArrayList<int>.

At runtime, all types are raw

  • Reason: type erasure
  • Something like if (object instanceof ArrayList<Dog>) will not compile because this check is impossible to execute at runtime.
  • The Class instances that you get are also always raw types. There is no ArrayList<Dog>.class, only ArrayList.class.

Type variables cannot be instantiated

  • If you have a type variable T, you cannot do new T(...) or new T[...] (array).
  • Reason: type erasure (you would be instantiating the erased value for T, not T itself).
  • If you want to construct objects of type T or arrays of type T inside a generic method, you will have to ask the caller for the right object or array constructor or for a Class object.
  • While you cannot instantiate an array of type T, you can easily create an ArrayList<T>. This is because ArrayList is a generic type itself, while in order to create an array of type T we would need the exact type T at runtime.

It’s impossible to create arrays of parameterized types

  • You can declare arrays of a parameterized type (e.g. AnimalWrapper<Dog>[])
  • You cannot instantiate an array of a parameterized type.
  • Reason: type erasure. At runtime, we would just get an AnimalWrapper[] array that allows any kind of AnimalWrapper without throwing an ArrayStoreException. If that is what you want, you can create an AnimalWrapper[] and then cast it to AnimalWrapper<Dog>[] (this will generate compiler warnings though).
  • The simplest solution is often to just create an ArrayList<AnimalWrapper<Dog>> instead.

Class type variables are not valid in static contexts

  • Type variables defined at the level of the class cannot be used in static contexts (static variables and static methods).
  • Example: if you have a class with type parameter T, you cannot have a static variable of type T.
  • Reason: type erasure. You can use a class multiple times with different values for T but a static variable only exists once (on the raw type), so it’s impossible to have a static variable with the exact type T for each of those values.
  • Remember that you can still use type variables in static contexts if they are not defined at the level of the class. For example, you can have a static method parameterized with type T if that type parameter is declared at the level of the method.

Methods may not clash after erasure

  • You are not allowed to declare methods that would clash after erasure (meaning that, after erasure, there would be two methods with the same signature).
  • This includes bridge methods! If you get a compiler error about methods clashing after erasure, it’s possible that the clash is generated by the bridge methods generated by the compiler. This is why it’s important to have some understanding of what these bridge methods are.

Exceptions: it is not possible to throw objects of a generic class.

  • Reason: type erasure. Catching instances of a generic class with a specific type parameter would require information that is not available at runtime.
  • It is still allowed to have a type variable in your throws declaration, as this is checked by the compiler.


  • Core Java SE 9 for the Impatient (book by Cay S. Horstmann)