How to create an instance for a given Type? [duplicate]

Question

With generics you can

var object = default(T); 

But when all you have is a Type instance I could only

constructor = type.GetConstructor(Type.EmptyTypes); var parameters = new object[0]; var obj = constructor.Invoke(parameters); 

or even

var obj = type.GetConstructor(Type.EmptyTypes).Invoke(new object[0]); 

Isn't there a shorter way, like the generics version?

Answer 1

The closest available is Activator.CreateInstance:

object o = Activator.CreateInstance(type); 

... but of course this relies on there being a public parameterless constructor. (Other overloads allow you to specify constructor arguments.)

I've used an explicitly typed variable here to make it clear that we really don't have a variable of the type itself... you can't write:

Type t = typeof(MemoryStream); // Won't compile MemoryStream ms = Activator.CreateInstance(t); 

for example. The compile-time type of the return value of CreateInstance is always object.

Note that default(T) won't create an instance of a reference type - it gives the default value for the type, which is a null reference for reference types. Compare that with CreateInstance which would actually create a new object.