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Home/ Questions/Q 701269
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T03:35:27+00:00 2026-05-14T03:35:27+00:00

I want to create a generic IEnumerable implementation, to make it easier to wrap

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I want to create a generic IEnumerable implementation, to make it easier to wrap some native C++ classes. When I try to create the implementation using a template parameter as the parameter to IEnumerable, I get an error.

Here’s a simple version of what I came up with that demonstrates my problem:

ref class A {};

template<class B>
ref class Test : public System::Collections::Generic::IEnumerable<B^> // error C3225...
{};

void test()
{
    Test<A> ^a = gcnew Test<A>();
}

On the indicated line, I get this error:

error C3225: generic type argument for ‘T’ cannot be ‘B ^’, it must be a value type or a handle to a reference type

If I use a different parent class, I don’t see the problem:

template<class P>
ref class Parent {};

ref class A {};

template<class B>
ref class Test : public Parent<B^> // no problem here
{};

void test()
{
    Test<A> ^a = gcnew Test<A>();
}

I can work around it by adding another template parameter to the implementation type:

ref class A {};

template<class B, class Enumerable>
ref class Test : public Enumerable
{};

void test()
{
    using namespace System::Collections::Generic;
    Test<A, IEnumerable<A^>> ^a = gcnew Test<A, IEnumerable<A^>>();
}

But this seems messy to me. Also, I’d just like to understand what’s going on here – why doesn’t the first way work?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T03:35:28+00:00Added an answer on May 14, 2026 at 3:35 am

    In your first example, your inheritance line should read:

    ref class Test : public System::Collections::Generic::IEnumerable<B>
    

    (no reference marker on the template)

    Then your usage line should read:

    Test<A^> ^a = gcnew Test<A^>();
    

    The reference markers go in the instantiation of the template, not the template itself.

    Here’s your sample, compilable:

    using namespace System;
    using namespace System::Collections::Generic;
    
    ref class A {};
    
    template<class B> ref class Test : public System::Collections::Generic::IEnumerable<B>
    {
    public:
        B GetInstance()
        {
            return Activator::CreateInstance<B>();
        }
    
        virtual System::Collections::IEnumerator^ GetEnumeratorObj() =
            System::Collections::IEnumerable::GetEnumerator
        {
            return nullptr;
        }
    
        virtual System::Collections::Generic::IEnumerator<B>^ GetEnumerator()
        {
            return nullptr;
        }
    };
    
    void test()
    {
        Test<A^> ^a = gcnew Test<A^>();
    }
    

    Edit: Realized I should explain why this is. To the best of my understanding, the reason that you can’t specify B^ in the IEnumerable inheritance is that IEnumerable is a generic with a constraint on it, while B is a template parameter which is unconstrained. Templates allow for much more flexible syntax, even when they govern ref objects, as they’re still effectively “parsed text” even in C++/CLI. However, when they bump into generics with constraints, the rules get a lot tighter.

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