Return Generic Type after determining Type Parameter dynamically

Question

I have a generic class as shown below

public class MyClass
{
    public T MyProp { get; set; }
}

Now I want to return the instance of

Answer 1

What you are trying to do is not possible. While you can create a generic type for an arbitrary generic type argument at runtime like this

public MyClass<object> ReturnWithDynamicParameterType(Type genericArgument)
{
    Type genericType = typeof(MyClass<>).MakeGenericType(genericArgument);
    return (MyClass<object>)Activator.CreateInstance(genericType);
}

this return line will always throw an InvalidCastException. As Lee already commented, classes are invariant. This means that for example MyClass<object> and MyClass<string> are simply not the same types. They are not even inherited from one another.

Even co-variance won't help since the out keyword is not allowed for generic parameters in classes.

I don't see a solution for this without knowing the types at compile-time. But I'm sure that you can solve what you are actually trying to achieve by other methods.

Answer 2

I am not sure if this fits your use, but your only way out is the use of an interface that is covariant.

public interface IMyClass<out T>
{
    T MyProp { get; }
}

public class MyClass<T> : IMyClass<T>
{
    public T MyProp { get; set; }
}

public IMyClass<object> ReturnWithDynamicParameterType()
{
    //This function determines the type of T at runtime and should return instance of MyClass<T>

    return new MyClass<string>();
}

This code compiles because your return type is not a class, but a covariant interface (note the out T on the type parameter). That interface allows retrieval only, so the get;set on the property has been replaced by a get on the interface.