Question

When having a function Foo(object) and an overload Foo(Exception), calls to Foo(null) are evaluated by Foo(Exception). Why is this?

UPDATE: (so basically most but not all nulls gets resolved to Foo(Exception))

Exception e = new Exception();

e = null;
Foo bar = new Foo(e); //Evaulated by Foo(Exception)
Foo bar = new Foo((object)e); //Evaluated by Foo(object)

object o = null;
Foo bar = new Foo(o); //Evaluated by Foo(object)
Foo bar = new Foo(null); //Evaulated by Foo(Exception)
Foo bar = new Foo((object)null); //Evaulated by Foo(object)

Thanks everyone.

Answer 1

The compiler resolves it at compile time by the type of reference of the parameter. For example:

Exception e = null;
Foo(e);

will be resolved to Foo(Exception e), in contrasts:

Exception e = null;
Foo((object)e);

will be resolved to Foo(object o). Note that the type of the instance is not taken into account.