Automatically making Base Constructors available in derived class?

I have a Base Class with two constructors, requiring a parameter:

public abstract class StoreBase 
{
    private readonly SomeObject_sobj;

    protected StoreBase(SomeObject sobj)
    {
        _sobj = sobj;
    }

    protected StoreBase(OtherObject oobj)
    {
        _sobj = new SomeObject(oobj);
    }
}

Then I have a derived class:

p开发者_如何学Pythonublic class MyDerived: StoreBase
{

}

This causes a compilation error as base class doesn't contain parameterless constructor.

My understanding is that because MyDerived doesn't contain a constructor, the compiler adds a parameterless constructor (that's well known and nothing to do with derived classes). However, as it derives from another class, the base class constructor needs to run first, and there is no way to determine which constructor should run from the empty MyDerived constructor.

Basically I'm asking: can I avoid copy/pasting all constructors from Base into Derived class if I really don't need additional constructor logic? Can I say "Take all constructors from base" without adding them all?

(And yes, I know I could/should refactor this into a parameterless constructor and a protected virtual Initialize() method. but I still wonder if I can work with constructors and still avoid copy/paste)

No - you will need to implement the (appropriate) constructors in the derived class, as well.

The derived class only needs to use one of the base constructors - so the constructors required in it may be completely different than the base class. They will need to be implemented by hand, even if that's just:

public class MyDerived : StoreBase
{
     public MyDerived(SomeObject sobj) : base(sobj) {}
     public MyDerived(OtherObject  oobj) : base(oobj) {}
}

Also:

(And yes, I know I could/should refactor this into a parameterless constructor and a protected virtual Initialize() method. but I still wonder if I can work with constructors and still avoid copy/paste)

Although I see this touted, I believe this is not always a good practice. In many cases, this is actually problematic, as you're relying on the subclass to properly call Initialize if they override your protected virtual method. For example, if the subclass did this, it could potentially be very bad:

public class MyDerived : StoreBase
{
   // .. other stuff
   protected override void Initialize()
   {
       // Leave out, intentionally or accidentally, the following:
       // base.Initialize(); 
   }
}

I actually avoid this in most situations, and initialize in the constructors (or in a private, non-virtual initialize method). Not doing this breaks any guarantees you have that your initialization will always occur the way you intend.

Constructors and constructor chaining provide the same functionality, with much better guarantees.

No, constructors are not inherited, and there is no shortcut for creating copies of the base constructors.

The closest is to implement constructors that call the base constructors. I guess you want to make them public anyway, so that you can actually create an instance of the class:

public MyDerived(SomeObject sobj) : base(sobj) {}
public MyDerived(OtherObject oobj) : base(oobj) {}

The compiler will only generate a default (parameterless) constructor if you haven't provided any constructor yourself. Once you define a constructor, the compiler won't generate one for you.

This can be important to keep in mind if you plan on serializing your class since serialization requires a default constructor. If your give your class a constructor other than a default one, then you'll also need to provide a default one to support serialization because, again, once you provide any constructor, the compiler won't provide one for you automatically.

Unfortunately the answer is no.

One thing you can do is call base from your MyDerived constructor:

class MyDerived: StoreBase
{
public MyDerived(OtherObject obj) : base (obj) {}
}

Can I say "Take all constructors from base" without adding them all?

No. :-(