How to Get class of Elements that List Contains?

Question

I have a list like that:

private List<T> myList = new ArrayList<T>();

I want to get the .class of T. How can I do that?

I mean as like:

myList.getClass()

EDIT: I tried that:

 Field genericField = Wrapper.class.getDeclaredField("myList");
 ParameterizedType genericType = (ParameterizedType) genericField.getGenericType();
 Class<?> genericClass = (Class<?>) genericType.getActualTypeArguments()[0];

and when I debug it

genericType

has a value of:

java.util.List

so I think that this is certainly different issue from that question: Get generic type of java.util.List because of if you declare a class with T and assign something to T later at another method, class that T holds disappear.

Answer 1

You can't because of type erasure. The generic type is not known at runtime (it's 'erased'), it's only used at compile time.

This is a major difference between java generics and c# generics for example.

Answer 2

Why do you need to know this? Perhaps this will do what you need

// must be at least one element (and all elements are the same)
Class elementType = myList.get(0).getClass(); 

---

The only way to do this is to either

• store the type in a field and add a getter method for this.

like

private final Class<T> type;
private final List<T> list = new ArrayList<T>();

public Class<T> getType() { return type; }

• use getGenericType() on the actual field, assuming the parameter type is NOT generic.
• use a sub class with a specific type.

like

List list = new ArrayList<Integer>() { }; // create a specific sub-class
final Class<? extends List> listClass = list.getClass();
final ParameterizedType genericSuperclass = (ParameterizedType) listClass.getGenericSuperclass();
Class elementType = (Class) genericSuperclass.getActualTypeArguments()[0];

This has the dis-advantage that you could end up creating lots of anonymous classes and potential confusion with things like ArrayList.class != list.getClass() however list instanceof ArrayList