Constructor with output-parameter

Question

today I saw a snipped that looked really horrible to me, but unfortunetly I cannot simply change it, so I wonder if I can bypass this somehow. I have a class with a constructor that has an output-parameter for success. But that looks really ugly to me. And now when deriving from this class I have to take this param with me- if I want to or not.

class ClassA {
    ClassA(out bool success) {...}
}

class B: ClassA {
    // call the constructor from ClassA but without the out-param
}

So I´d know if its good practise or if not how I can avoid declaring the out-param from ClassB.

Answer 1

Well, the design of that class is broken anyway, so let's break it a bit more (NOTE! I do not recommend this approach!):

void Main()
{

}

public class ClassA
{
    public ClassA(out bool success)
    {
        success = true;
    }
}

public class B: ClassA
{
    private static bool success;

    // call the constructor from ClassA but without the out-param
    public B()
        : base(out success)
    {
    }
}

Other than that, the closest you can get is making a factory method:

public class B : ClassA
{
    public static B Create()
    {
        bool success;
        var result = new B(out success);
        if (success)
            return result;
        // TODO: Dispose result?
        throw new StupidProgrammerException();
    }
}