Derived class: using Base class member in initializer list

Question

First snippet of code:

struct Base
{
    int x{};
};

struct Derived :
    Base
{
    Derived()
        : y{x}
    {
    }

    int y;
};

int main()
{
    Derived d;
}

Compiles fine on:

• gcc (6.0.0)
• clang (3.8.0)
• vc++ (Visual Studio 2013 Update 4, 18.00.31101)

Second snippet of code:

#include <type_traits>

template<int N>
struct Base
{
    int x = N;
};

static const int When0 = -1;

template<int N>
struct Derived :
    std::conditional<N == 0,
        Base<When0>,
        Base<N>>::type
{
    Derived()
        : y{x}
    {
    }

    int y;
};

int main()
{
    Derived<0> d;
}

Compiles fine on:

• vc++

Will not compile on:

• gcc
• clang

To fix gcc and clang, I need to specify x 's class:

#include <type_traits>

template<int N>
struct Base
{
    int x = N;
};

static const int When0 = -1;

template<int N>
struct Derived :
    std::conditional<N == 0,
        Base<When0>,
        Base<N>>::type
{
    using base_t = typename std::conditional<N == 0,
        Base<When0>,
        Base<N>>::type;

    Derived()
        : y{base_t::x}
    {
    }

    int y;
};

int main()
{
    Derived<0> d;
}

See (vc will compile it too):

• clang
• gcc

Questions: Which compiler(s) is correct ? What standard says about this ?

Thanks

Answer 1

This is the standard problem of accessing a base class (non-dependent) member from a templated derived class. See this FAQ entry .

Changing it to simply this->x also works, so VC++ is wrong here.