How to create instance of the unkown type?

Question

I have a couple of functions which requires exact argument type (aka T):

private  void doWork1(T _obj) {...}
private  void doWork2         


        

Answer 1

User the type Number for your obj. Integer and Double both extend this type.

The abstract class {@code Number} is the superclass of platform classes representing numeric values that are convertible to the primitive types {@code byte}, {@code double}, {@code float}, {@code int}, {@code long}, and {@code short}.

public void parse(int id)
{
    Number obj = null;
    switch (id)
    {
        case 1:
        {
            obj = new Integer(1);
            break;
        }
        case 2:
        {
            obj = new Double(2);
            break;
        }
    }
    doWork1(obj);
    doWork2(obj);
    doWork3(obj);
}

If you do not want to be this concrete, you can always use Object.

Answer 2

You can use Number or Object, which are both common supertypes of Integer and Double.

---

However, the generics are unnecessary:

private <T> void doWork1(T _obj) {...}

is identical to

private void doWork1(Object _obj) {...}

after erasure.

The only point of having a type variable for an input parameter is if:

• You need to indicate that the generics of another input parameter need to be related, e.g. you are passing T _obj and List<T> _list.

  Note that you don't need a generic type for T _obj1 and T _obj2, though - that degenerates to the upper bound of T (e.g. Object);

• If you need it to be related to the return type:

  <T> T doWork1(T _obj) { ... }
  

You don't need either case here, so just remove the unnecessary complication.