I can not understand why the constructor is executed with the parameter Double[]
?
using System.Collections.Generic; using System.Linq; using System.Text; namespace MyConsoleApp { class Program { static void Main(string[] args) { D myD = new D(null); Console.ReadLine(); } } public class D { public D(object o) { Console.WriteLine("Object"); } public D(double[] array) { Console.WriteLine("Array"); } public D(int i) { Console.WriteLine("Int"); } } }
I think because the first constructor takes a parameter of reference type. The first constructor with a reference parameter because null
is the default value for reference types.
But i don't understand why not object
, it's also a reference type.
But I can not understand why no object? It's also a reference type?
Yes, both double[]
and object
are reference types, so null
is implicitly convertible to both of them. However, member overloading generally favours more specific types, so the double[]
constructor is used. See section 7.5.3 of the C# specification for more details (and boy are there a lot of details).
In particular, from section 7.5.3.5:
Given two different types T1 and T2, T1 is a better conversion target than T2 if at least one of the following holds:
- An implicit conversion from T1 to T2 exists, and no implicit conversion from T2 to T1 exists
That's the case here, where T1
is double[]
and T2
is object
. There's an implicit conversion from double[]
to object
, but no implicit conversion from object
to double[]
, so double[]
is a better conversion target than object
.
If you want force the use of the object
constructor, just cast:
D myD = new D((object) null);
Basically, double[]
is an object
, but all object
s are not double[]
s. As double[]
's the more specific option, the compiler chooses it, as the most specific one.
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