Question

Consider the following code:

The code

public class RecursiveConstructor
{
   //When this constructor is called 
   public RecursiveConstructor():this(One(), Two())
   {
       Console.WriteLine("Constructor one. Basic.");
   }

   public RecursiveConstructor(int i, int j)
   {
       Console.WriteLine("Constructor two.");
       Console.WriteLine("Total = " + (i+j));
   }

   public static int One()
   {
       return 1;
   }

   public static int Two()
   {
       return 2;
   }
}

The calling method

public class RecursiveConstructorTest
{
    public static void Main()
    {
        RecursiveConstructor recursiveConstructor = new RecursiveConstructor();

        Console.ReadKey();
    }
}

The Result

Constructor two.

Total = 3

Constructor one. Basic.

Why is the 2nd constructor run first?

I understand that in chained constructors we call the base class constructor first and then make our way back up the chain but when the constructor is held in the same class why do we still see this behaviour where the extra constructor is called first?

I would have thought that the most basic constructor contents would be executed first.

Answer 1

I think the compiler runs the safer scenario.
If you call another constructor here, there are chances that this other constructor is a prerequisite to your current constructor. This behaviour is consistent with the one exposed when calling base constructors, and is then to be expected.

When creating a new instance of a class, there is a chain of constructors that get called from the least specialized (the constructor of the object class) to the most specialized (the constructor of your current class).

The operator : allows you to explicitly add a constructor to this chain, so this order seems natural.