Workaround to template specialization

Question

Abstract Base Class

abstract class Base { public abstract void Method<T>(T args) { } }

Derived Class several of these exist, each with a replacement for T

class Derived : Base { public override void Method<int>(int args) { } }

Now, I know this isn't supported, but I need it because somewhere else in my code

class AnotherBaseClass<T, E> where E : Base { E e; // Will be actually one of the derived classes T t; // Simple types like string or int // I want this to work basically public void Func() { e.Method(t); } }

this is going on.

What is the easiest solution to this? If I am able to do this, it will save me from writing a lot of code.

Eldar S · Accepted Answer · 2021-09-02 21:18:54Z

2

If you're fine with making Base a generic class, this is how:

abstract class Base<T> { public abstract void Method(T args); } class Derived : Base<int> { public override void Method(int args) {} } class AnotherBaseClass<T, E> where E : Base<T> { E e; T t; public void Func() { e.Method(t); } }

If you need Base to not be generic, you can add this:

abstract class Base { public void Method<T>(T args) { var genericSelf = this as Base<T>; genericSelf.Method(args); } }

and make Base<T> inherit Base and ensure your concrete classes always derive Base<T> and never Base directly.

edited Sep 2, 2021 at 21:18

answered Sep 2, 2021 at 19:17

Eldar S

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4 Comments

demberto Over a year ago

Yeah, I forgot to write that I tried the first method (making Base a generic class), but it doesn't work when I want to use Base as a function argument for e.g. public void Fn(Base base). With the 2nd method I can use Base for function arguments and Base<T> for subclasses, right?

Eldar S Over a year ago

You must subclass Base<T> for concrete classes as I said, don't derive Base`` directly or else the this as Base<T>` will return null.

demberto Over a year ago

Yes I am using Base<T> for all concrete sealed classes and Base is just used for the function signature as I could not do public void Fn(Base<T> base). It gave compile errors.

Eldar S Over a year ago

you need to write public void Fn<T>(Base<T> base) for it to compile

MikeJ · Accepted Answer · 2021-09-02 20:21:15Z

I'm not exactly sure what problem you're sovling but under .Net 5 what you have in your question works as is...

abstract class Base { public abstract void Method<T>(T args); } class Derived : Base { public override void Method<T>(T args) { Console.Write($"{nameof(Derived)}.Method<{typeof(T)}>({args})"); } } class AnotherBaseClass<T, E> where E : Base { readonly E e; readonly T t; public AnotherBaseClass(T t, E e) { this.e = e; this.t = t; } public void Func() { e.Method(t); } }

Run like so:

 static void Main(string[] args) { Derived d = new Derived(); var another = new AnotherBaseClass<int,Derived>(5, d); another.Func(); }

Produces: "Derived.Method<System.Int32>(5)"

The question was about having a derived class with a non-generic method but rather a method implemented for a specific type, e.g. int
I am using .NET Framework 4.5, and yeah as @EldarS mentioned, this is not the solution I wanted
@EldarS this is literally what he said he was trying to do in his question. The code is not mine, it's his.

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