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Home/ Questions/Q 4084548
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T18:27:15+00:00 2026-05-20T18:27:15+00:00

I am working on a small project and I came across that problem. The

  • 0

I am working on a small project and I came across that problem.

The project output is a library containing an interface. I would like to implement that interface and seal the functions in it like this if possible:

public interface ITest
{
    void SomeMethod();
}

class A : ITest
{
    public sealed override SomeMethod()
    {

    }
}

The idea is to have the interface available to everyone and have some specialized class that implements it. The exception is that I want to make sure that if someone create a specialized class of type A, he/she won’t be able to change the method’s behavior.

The problem is you can’t put the “override” keyword in there since the method isn’t declared as “virtual” in the interface. And you can’t declare it as “virtual” in the interface since it’s not allowed. And you can’t remove the “override” keyword since it’s needed by “sealed”.

Any workaround or brainstorming idea would be welcome, but if someone can come up with a solution that includes an interface, I’d be really happy to learn it!

Thanks!

EDIT: Forget this question! Like Ani said, I forgot that by default method in C# are sealed. Seems like it’s always good to go back to the basics once in a while…

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  1. Editorial Team
    Editorial Team
    2026-05-20T18:27:15+00:00Added an answer on May 20, 2026 at 6:27 pm

    I may have completely misunderstood the question, but if your intention is to seal the method in A, you can just do:

    class A : ITest
    {
        public void SomeMethod()  { ... }
    }
    

    Unlike Java, methods in C# are sealed by default. Subclasses of A won’t be able to override the method since it hasn’t been marked virtual.

    On the other hand, if your intention is to mark the method ‘almost sealed’ in the interface, so that it forces upon an implementing class to immediately seal it, that isn’t possible. It isn’t (and shouldn’t be) the business of the interface to dictate such details of implementation – an interface is meant to represent a specification.

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