Guava TypeToken and Common Classes

Question

I use the Guava TypeToken class in my project, but I get an unexpected result.

I have MyGenericClass<T> :

 public class MyGenericClass<T> implements MyInterface { private TypeToken<T> recordType; public MyGenericClass(String name) { this.recordType = new TypeToken<T>(getClass()) {}; // ... } // ... @SuppressWarnings("unchecked") protected Class<T> getRecordType() { return (Class<T>) recordType.getRawType(); } } 

So, if I create an object via new MyGenericClass<String>() and then call getRecordType() , I expect to get java.lang.String , instead I get java.lang.Object .

But, if I extend the general class:

 public class MyStringImpl extends MyGenericClass<String> { // ... } 

and create an instance of this new class: new MyStringImpl() , then I get the correct result.

Why is this happening? Is this the expected TypeToken behavior?

Answer 1

To do this, you will need the same anonymous subclass trick when you instantiate MyGenericClass :

 new MyGenericClass<String>() {} 

If you do this, you will get the String returned by getRecordType . The reason why this is necessary is explained in JavaDocs for this TypeToken constructor .

Clients create an empty anonymous subclass. At the same time, this parameter is introduced into the hierarchy of types of the anonymous type, so we can restore it at runtime, despite the erasure.

Answer 2

To add some boring details to Yang's answer: It would be nice if the TypeToken worked the way you expected, but that is not possible. When you announce

 public class MyGenericClass<T> implements MyInterface {...} 

JVM sees something like

 public class MyGenericClass<Object> implements MyInterface {...} 

due to erasure.

But when you announce

 public class MyStringImpl extends MyGenericClass<String> {...} 

then in the definition of MyStringImpl generics used are written and can be obtained using Class#getGenericSuperclass() . This is (part of) the magic behind TypeToken .