Convert java.lang.reflect.Type to Class clazz

Question

How can I convert java.lang.reflect.Type to Class clazz?

If I have one method as next which has an argument of Class

Answer 1

If you are willing to use a library, you could use com.google.guava:guava:12+:

Class<?> clazz = com.google.common.reflect.TypeToken.of(type).getRawType();

Alternatively you could also use com.fasterxml.jackson.core:jackson-databind:2.8.x:

Class<?> clazz = com.fasterxml.jackson.databind.type.TypeFactory.rawClass(type);

This handles all cases correctly and you will get the type-erased class of your type.

Answer 2

Did you mean this?

public <T extends Type> void oneMethod(T clazz) {

}

public void someMethod(Type type) {
    oneMethod(type);
}

Answer 3

You have to ensure that type is an instance of Class, and then cast it.

if (type instanceof Class) {
  Class<?> clazz = (Class<?>) type;
  otherMethod(clazz);
}

Of course, you also have to handle the case of it not being a Class.

Answer 4

Andy Turner answer is correct, however if you need to retrieve a class from a type parameter this is a complete example:

private static class MyClass extends ArrayList<Integer> {
}

public static void main(String[] args) {

    ParameterizedType arrayListWithParamType
            = (ParameterizedType) MyClass.class.getGenericSuperclass();

    Type integerType = arrayListWithParamType.getActualTypeArguments()[0];

    Class<?> integerClass = (Class<?>) integerType;

    System.out.println(integerClass == Integer.class);
}

Answer 5

It would be weird that a Type would be anything else than a Class... Javadoc for Type says

All Known Implementing Classes: Class

So unless you have special libraries that use non Class Types, you can simply cast - but you must be ready for a possible ClassCastException. Beware: Java use undocumented Type implementation to represent generics, see below.

You can explicitely process it or not because it is an unchecked exception:

Explicit way:

try {
    Class<?> clazz = (Class<?>) type;
}
catch (ClassCastException ex) {
    // process exception
}

Implicit way:

Class<?> clazz = (Class<?>) type;

but the current method could throw...

---

EDIT per @Andy Turner's comment:

Beware: Type type = new ArrayList<String>().getClass().getGenericSuperclass(); yields something that's a Type but not a Class. This one is a ParameterizedType, so you can use getRawType() method to find the actual class, but others might exist.

Answer 6

Using generic types in runtime is a little bit tricky in Java. And I think this is a root cause of your issue.

1) to be sure about generic in runtime we doing like this:

class MyClass<E> {}

and then:

MyClass<TargetType> genericAwaredMyClassInctance = new MyClass<TargetType>(){};

please pay attention to {} in the end. It means anonymous extends of MyClass. This is an important nuance.

2) let`s improve MyClass to be able to extract the type in runtime.

class MyClass<E> {

    @SuppressWarnings("unchecked")
    protected Class<E> getGenericClass() throws ClassNotFoundException {
        Type mySuperclass = getClass().getGenericSuperclass();
        Type tType = ((ParameterizedType)mySuperclass).getActualTypeArguments()[0];
        String className = tType.getTypeName();

        return (Class<E>) Class.forName(className);
    }

}

and finally, use it like this:

MyClass<TargetType> genericAwaredMyClassInctance = new MyClass<TargetType>(){};

assert(genericAwaredMyClassInctance.getGenericClass() == TargetType.class)