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Home/ Questions/Q 8848929
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T12:31:11+00:00 2026-06-14T12:31:11+00:00

I am quite sure I have read the reason why the compiler cannot cope

  • 0

I am quite sure I have read the reason why the compiler cannot cope with this code somewhere on SO, but, after a few hours of searching, I still can’t find it. Here is the relevant code:

#include <iostream>

template <typename T>
class base
{
};

class derived : base<derived::myStruct>
{
public:
    struct myStruct
    {
    };
};

int main ()
{
    return 0;
}

The problem is that the parser first tries to generate the base<derived::myStruct> specialization before parsing derived, and, thus, I get this error: “error C2065: ‘myStruct’ : undeclared identifier”. As a silly trick, I noticed that VS2010 stops complaining if I pre-declare struct myStruct; just above class derived. In my opinion, myStruct should be bound inside derived and this code should throw the same error:

#include <iostream>

template <typename T>
class base
{
};

struct myStruct;

class derived : base<derived::myStruct>
{
public:
    struct myStruct
    {
    };
};

int main ()
{
    return 0;
}

Update: gcc-4.5.1 is able to throw the expected error, so, I guess the above is a bug in VS2010…

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  1. Editorial Team
    Editorial Team
    2026-06-14T12:31:12+00:00Added an answer on June 14, 2026 at 12:31 pm

    A workaround is to declare myStruct outside of derived.

    #include <iostream>
    
    template <typename T>
    class base
    {
    };
    
    struct derived_myStruct
    {
    };
    
    class derived : base<derived_myStruct>
    {
    public:
        typedef derived_myStruct myStruct;
    };
    
    int main ()
    {
        return 0;
    }
    

    In more complex cases where derived is a template class, derived_myStruct would too be a template class, with the same template paramerters (or just a subset), and you would pass those through.

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