Creating an instance of variable with reflection

Question

I want to create an instance of type t with reflection, that is


Type t = typeof(string);
string s = (t)Activator.CreateInstance(t); // this fails because of convertion
string s = Activator.CreateInstance(t) as t // also fails

Is there a way to perform such a convertion? Thanks.

Why would you want to do this? If you know the compile-time type of the variable then why not construct the object normally? — Eric Lippert
– Eric Lippert, Commented Jan 6, 2010 at 15:24

David M · Accepted Answer · 2010-01-06 10:58:59Z

4

Yes. You have to convert to string, not to t. You may want a generic method, alternatively:

public T GetInstance<T>()
{
    Type t = typeof(T);
    T s = (T)Activator.CreateIstance(t);
    return s;
}

As things stand you are attempting to cast an object that is actually an instance of System.String to type System.Type...

answered Jan 6, 2010 at 10:58

David M

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3 Comments

LukeH Over a year ago

You could just add a where T : new() contstraint on the method and the body would then be a single line: return new T();

David M Over a year ago

Very true, and would be safer as types not implementing a paramaterless constructor could be picked up at compile time.

Jaco Pretorius Over a year ago

If I recall correctly there is a generic overload as well, so you should be able to do Activator.CreateInstance<T>();

Rune Grimstad · Accepted Answer · 2010-01-06 10:58:27Z

1

Try this:

string s = (string)Activator.CreateInstance(t);

Activator.CreateInstance returns an instance boxed in an object so it must be cast to the correct type before you can use it.

In your example t is a Type object variable and not a type reference. You must either specify the type directly as in my example or you can use generics.

answered Jan 6, 2010 at 10:58

Rune Grimstad

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