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Home/ Questions/Q 4094600
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Editorial Team
  • 0
Editorial Team
Asked: May 20, 20262026-05-20T19:45:55+00:00 2026-05-20T19:45:55+00:00

I wonder why IEnumerable<int> can’t be assigned to a IEnumerable<object> . After all IEnumerable

  • 0

I wonder why IEnumerable<int> can’t be assigned to a IEnumerable<object>. After all IEnumerable is one of the few interfaces that supports covariance…

  • The subtype relation and covariance stuff works with reference types
  • int seems to be a proper subtype of object

The combination of both features doesn’t work however…

class A
{
}

class B : A
{
}

class Program
{
    static void Main(string[] args)
    {
        bool b;
        b = typeof(IEnumerable<A>).IsAssignableFrom(typeof(List<B>));
        Console.WriteLine("ienumerable of ref types is covariant: " + b); //true

        b = typeof(IEnumerable<object>).IsAssignableFrom(typeof(List<int>));
        Console.WriteLine("ienumerable of value tpyes is covariant: " + b); //false

        b = typeof(object).IsAssignableFrom(typeof(int));
        Console.WriteLine("int is a subtype of object: " + b); //true
    }
}

thanks for your help!
sebastian

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  1. Editorial Team
    Editorial Team
    2026-05-20T19:45:56+00:00Added an answer on May 20, 2026 at 7:45 pm

    Value types aren’t LSP-subtypes of object until they’re boxed.

    Variance doesn’t work with value types. At all.


    Demonstration that int is not a proper subtype (subtype in the LSP sense) of object:

    Works:

    object x = new object();
    lock (x) { ... }
    

    Does not work (substitutability violated):

    int y = new int();
    lock (y) { ... }
    

    Returns true:

    object x = new object();
    object a = x;
    object b = x;
    return ReferenceEquals(a, b);
    

    Returns false (substitutability violated):

    int y = new int();
    object a = y;
    object b = y;
    return ReferenceEquals(a, b);
    

    Of course, the topic of the question (interface variance) is a third demonstration.

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