How does Gson TypeToken work?

问题:

I understand that in Java contrary to, for example, C# generics are compile-time feature and is removed via type erasure. So, how does Gson's TypeToken really work? How does it get the generic type of an object?

回答1:

From §4.6 of the JLS (emphasis mine):

Type erasure is a mapping from types (possibly including parameterized types and type variables) to types (that are never parameterized types or type variables). We write |T| for the erasure of type T. The erasure mapping is defined as follows:

The erasure of a parameterized type (§4.5) G is |G|.

The erasure of a nested type T.C is |T|.C.

The erasure of an array type T[] is |T|[].

The erasure of a type variable (§4.4) is the erasure of its leftmost bound.

The erasure of every other type is the type itself.

Therefore, if you declare a class with an anonymous subclass of itself, it keeps it's parameterized type; it's not erased. Therefore, consider the following code:

import java.lang.reflect.ParameterizedType;
import java.util.Arrays;
import java.util.HashMap;

public class Erasure<T>
{
    public static void main(String...strings) {
      Class<?> foo = new Erasure<HashMap<Integer, String>>() {}.getClass();
      ParameterizedType t = (ParameterizedType) foo.getGenericSuperclass();
      System.out.println(t.getOwnerType());
      System.out.println(t.getRawType());
      System.out.println(Arrays.toString(t.getActualTypeArguments()));
    }
}

This outputs:

null
class Erasure
[java.util.HashMap<java.lang.Integer, java.lang.String>]

Notice that you would get a ClassCastException if you did not declare the class anonymously, because of erasure; the superclass would not be a parameterized type, it would be an Object.

回答2:

Java's type erasure applies to individual objects, not classes or fields or methods. TypeToken uses an anonymous class to ensure it keeps generic type information, instead of just creating an object.