Question

Assume you have this:

// General purpose
public interface ISerializer
{
    IDataResult Serialize<T>(T instance);
}

// General purpose
public interface IDataResult
{
}

// Specific - and I implement IDataResult
public interface IMyCrazyDataResult : IDataResult
{
}

public class MyCrazySerializer : ISerializer
{
    // COMPILE ERROR:
    // error CS0738: 'MyCrazySerializer' does not implement interface member 'ISerializer.Serialize<T>(T)'. 
    // 'MyCrazySerializer.Serialize<T>(T)' cannot implement 'ISerializer.Serialize<T>(T)' because it does 
    // not have the matching return type of 'IDataResult'.
    public IMyCrazyDataResult Serialize<T>(T instance)
    {
        throw new NotImplementedException();
    }
}

Why in the world do I get this compile error? I am respecting the interface – I do, in-fact, return an IDataResult, albeit indirectly. Is it that the compiler can’t figure that out or is there something fundamentally (at an OO level) wrong, here?

I thought the entire point of having an interface was that I could guarantee some implementation, but leave it open for me to add-on to it. That is what I am doing – yet I get a compile error.

In my real code, I want the return type to be a bit more specific because I have several additional methods that I have in my derived interface. If I make the return type of MyCrazySerializer.Serialize of-type IDataResult, then intellisense just shows me and the bare-bones common methods, where I want to show a more-specific interface.

How else could I accomplish this? What is wrong with this code???

Answer 1

C# does not support return type covariance so you’ll need to implement the Serialize<T> method exactly as it appears on the interface. You could implement it explicitly however, meaning any clients which know the real type of MyCrazySerializer can access the more specific method:

public class MyCrazySerializer : ISerializer
{
    public IMyCrazyDataResult Serialize<T>(T instance)
    {
        throw new NotImplementedException();
    }

    IDataResult ISerializer.Serialize<T>(T instance)
    {
        return this.Serialize(instance);
    }
}

As the comment point out, you can simply call the more specific version in your explicit implementation.

Which you can use as:

IMyCrazyDataResult result = new MyCrazySerializer().Serialize<int>(1);
ISerializer serializer = (ISerializer)new MyCrazySerializer();
IDataResult = serializer.Serialize<int>(1);