Question

Let’s say I have 2 constructors.

public class MyClass
{
   public MyClass()
   {
     int id = 0;
     //Same behaviour
   }
   Public MyClass(int id)
   {
     //Same behaviour
   }
}

Both constructions implement the same behavior. The only difference is that, if the first constructor is called and the value of id = 0;

My question is to know if I can call the second constructor, instead of implemetanting the same behavior? If that’s possible, do I do it?

Answer 1

Yes, this is called constructor chaining. It’s achieved like so:

public class MyClass {
    public MyClass() : this(0) { }
    public MyClass(int id) {
        this.id = id;
    }
}

Note that you can chain to the base-class constructor like so:

public class MyClass : MyBaseClass {
    public MyClass() : this(0) { }
    public MyClass(int id) : base(id) { }
}

public class MyBaseClass {
    public MyBaseClass(int id) {
        this.id = id;
    }
}

If there is a base class and you don’t specify a constructor to chain to, the default is the accessible parameterless constructor, if there is one. If you do not specify a constructor to chain to and there is no accessible parameterless constructor, you will get a compile-time error.