Question

Reading existing code at work, I wondered how come this could work. I have a class defined in an assembly :

[Serializable]
public class A
{
    private readonly string _name;
    private A(string name)
    {
        _name = name;
    }
}

And in another assembly :

public void f(Type t) {
    object o = Activator.CreateInstance(t);
}

and that simple call f(typeof(A))

I expected an exception about the lack of a parameterless constructor because AFAIK, if a ctor is declared, the compiler isn’t supposed to generate the default public parameterless constructor.

This code runs under .NET 2.0.

[EDIT] I’m sorry but I misread the actual code… The sample I provided doesn’t illustrate it. I accepted JonH answer because it provided a good piece of information.

Answer 1

See this: Creating instance of type without default constructor in C# using reflection

Here’s to the future also, this is in regards to C# 4.0:

Posted by Microsoft on 1/4/2010 at
2:08 PM
We have reviewed your bug and
have determined that the behavior that
you described is by design. We are now
archiving this issue. Thanks for using
Visual Studio and the .Net Framework!

There are many overloads of Activator.CreateInstance, the one that
takes a single type parameter only
invokes the default parameterless
constructor. Constructors that take an
optional parameter like the one in
your example are not default
constructors. To invoke them you need
to:

1. use an overload that takes an argument array
2. Pass in Type.Missing as the argument
3. Specify OptionalParamBinding in the BindingFlags

Here is an example:

Activator.CreateInstance(typeof(MyClassName),
 BindingFlags.CreateInstance |
 BindingFlags.Public |
 BindingFlags.Instance |
 BindingFlags.OptionalParamBinding,
 null, new Object[] {Type.Missing}, null);

Thanks,

Weitao Su

Microsoft Corp.