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Home/ Questions/Q 746023
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T09:39:04+00:00 2026-05-14T09:39:04+00:00

Having defined this interface: public interface IInputBoxService<out T> { bool ShowDialog(); T Result {

  • 0

Having defined this interface:

public interface IInputBoxService<out T> {
    bool ShowDialog();
    T Result { get; }
}

Why does the following code work:

public class StringInputBoxService : IInputBoxService<string> {
    ...
}

...

IInputBoxService<object> service = new StringInputBoxService();

and this doesn’t?:

public class IntegerInputBoxService : IInputBoxService<int> {
    ...
}

...

IInputBoxService<object> service = new IntegerInputBoxService();

Does it have anything to do with int being a value type? If yes, how can I circumvent this situation?

Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-14T09:39:05+00:00Added an answer on May 14, 2026 at 9:39 am

    Yes, it absolutely has to do with int being a value type. Generic variance in C# 4 only works with reference types. This is primarily because references always have the same representation: a reference is just a reference, so the CLR can use the same bits for something it knows is a string reference as for an object reference. The CLR can make sure that the code will be safe, and use native code which only knows about IInputBoxService<object> when passed an IInputBoxService<string> – the value returned from Result will be representationally compatible (if such a term exists!).

    With int => object there would have to be boxing etc, so you don’t end up with the same code – that basically messes up variance.

    EDIT: The C# 4.0 spec says this in section 13.1.3.2:

    The purpose of variance annotations is
    to provide for more lenient (but still
    type safe) conversions to interface
    and delegate types. To this end the
    definitions of implicit (§6.1) and
    explicit conversions (§6.2) make use
    of the notion of
    variance-convertibility, which is
    defined as follows: A type T is variance-convertible to a type
    T if T is either an
    interface or a delegate type declared
    with the variant type parameters T, and for each variant type
    parameter Xi one of the following
    holds:

    • Xi is covariant and an
      implicit reference or identity
      conversion exists from Ai to Bi

    • Xi
      is contravariant and an implicit
      reference or identity conversion
      exists from Bi to Ai

    • Xi is invariant
      and an identity conversion exists from
      Ai to Bi

    This doesn’t make it terribly obvious, but basically reference conversions only exist between reference types, which leaves only identity conversions (i.e. from a type to itself).

    As for workarounds: I think you’d have to create your own wrapper class, basically. This can be as simple as:

    public class Wrapper<T>
    {
        public T Value { get; private set; }
        public Wrapper(T value)
        {
            Value = value;
        }
    }
    

    It’s pretty nasty though 🙁

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