Initializing object of derived class using initializer-list

Question

I got a struct B that is derived from a struct A.

struct A{
  int a;  
};

struct B : public A{
    int b;
};

Is there a straight way to initialize an object of type B without providing a constructor, let's say using initializer-list?

Some more insight:

I got two structs that I use to pass data between threads; the second one holds the same data as the first one, with the addition of some synchronization variable. I could make the first struct a data member of the second one, or just duplicate the declaration of data members in the second struct, to easily use the initializer-list; but I think that in this particular application it would be logically more correct that the second struct extends the first one.

Answer 1

There isn't an extremely concise solution, but there is a solution at least:

#include <type_traits>
#include <utility>

struct B : A
{
    int b;

    template <typename ...Args,
              typename = typename std::enable_if<
                             std::is_constructible<A, Args&&...>::value>
    B(int x, Args &&... args)
    : b(x), A(std::forward<Args>(args)...)
    { }
};

This solution isn't exactly short, but it is general. The new constructor for B exists only in those specializations which make sense, thanks to the enable-if SFINAE, so B is precisely as constructible as it can be.

There is one more danger here that I didn't address, namely how explicit the new constructor should be. Ideally it should be as explicit as the matching A constructor, but that's a bit hard to detect programatically (as done in N4064 for pairs and tuples).