当T实现类时,C++/CLI中的IComparable <T>

Question

更新:如评论中所述,C++/CLI应该是value struct; 编译器错误'明确'陈述"必须是值类型".

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在C#中,我可以写

      public struct Id<T> : IComparable<Id<T>>
    {
        public int CompareTo(Id<T> other)
        {
            throw new NotImplementedException();
        }
    }

当我尝试在C++/CLI中执行相同操作时

generic<typename T>
public ref struct Id : System::IComparable<Id<T>>
{
public:
    virtual int CompareTo(Id<T> other)
    {
        throw gcnew System::NotImplementedException();
    }
};

我收到编译器错误error C3225: generic type argument for 'T' cannot be '...::Id<T>', it must be a value type or a handle to a reference type.

难道这仍然是不固定的编译器错误？

Answer 1

在C++/CLI中,您需要在托管类型上使用句柄.这编译:

generic<typename T>
public ref struct Id : System::IComparable<Id<T>^>
{
public:
    virtual int CompareTo(Id<T>^ other)
    {
        throw gcnew System::NotImplementedException();
    }
};