Is raw type of ClassName equivalent to ClassName

Question

The title is pretty self-explanatory. Will the raw type of, for example, ArrayList act the same as ArrayList<Object>?

Define "act the same". Do you mean run-time? Then read about [type erasure](http://docs.oracle.com/javase/tutorial/java/generics/erasure.html). — PM 77-1, Nov 10 '13 at 17:22
Generics is a compile time feature. At runtime it makes little difference. — Peter Lawrey, Nov 10 '13 at 17:51

score 3 · Accepted Answer · edited May 23 '17 at 10:25

Not quite. Take a look at

  MyClass myClass1 = new MyClass<Integer>();
//MyClass<Object> myClass2 = new MyClass<Integer>();//compilation error
  MyClass<?> myClass3 = new MyClass<Integer>();//this time it compiles

  myClass1.setter(1);
//myClass3.setter(1);//compilation error

In case of myClass2 it will not compile because you are declaring reference to holder of Objects but you are passing holder of Integers and generics are not covariant

Also if you change <Object> to <?> wildcard like in case of myClass3 you wouldn't be able to use its setter with any object (beside null) like myClass3.someSetter(1) because MyClass<?> can be reference to MyClass<Anything> that can hold any type of objects, and passing 1 to holder of Strings would not be safe.

ZhongYu · Answer 2 · 2013-11-10T17:40:16.947

In terms of method signatures, the erased E is its bound Object, so it is ok to say that the raw ArrayList has member methods similar to ArrayList<Object>

In terms of subtyping, raw type is more like wildcard type; ArrayList is a super type of ArrayList<x> for any x. On the other hand, ArrayList<x> can be converted to raw ArrayList without raising any compiler error/warning.

This is all for backward compatibility. You should avoid raw types anyway so you don't need to know these useless information.

2 Answers2