Hide empty base class for aggregate initialization

Question

Consider the following code:

struct A
{
    // No data members
    //...
};

template<typename T, size_t N>
struct B : A
{
    T data[N];
}

This is how you have to initialize B: B<int, 3> b = { {}, {1, 2, 3} }; I want to avoid the unnecessary empty {} for the base class. There is a solution proposed by Jarod42 here, however, it doesn't work with elements default initialization: B<int, 3> b = {1, 2, 3}; is fine but B<int, 3> b = {1}; is not: b.data[1] and b.data[2] aren't default initialized to 0, and a compiler error occurs. Is there any way (or there will be with c++20) to "hide" base class from construction?

Answer 1

The easiest solution is to add a variadic constructor:

struct A { };

template<typename T, std::size_t N>
struct B : A {
    template<class... Ts, typename = std::enable_if_t<
        (std::is_convertible_v<Ts, T> && ...)>>
    B(Ts&&... args) : data{std::forward<Ts>(args)...} {}

    T data[N];
};

void foo() {
    B<int, 3> b1 = {1, 2, 3};
    B<int, 3> b2 = {1};
}

If you provide fewer elements in the {...} initializer list than N, the remaining elements in the array data will be value-initialized as by T().