C ++ 14标准布局类型可以使用`alignas`作为字段吗？

Question

I'd like to use templates to simplify the construction of unions with non-trivial types. The following seems to "work" in practice, but is technically not legal by the spec:

template<typename T> struct union_entry {
  void (*destructor_)(void *);  // how to destroy type T when active
  T value_;
};
union U {
  union_entry<A> a;
  union_entry<B> b;
  // ... some constructor and destructor...
};

The problem is that (according to N4141) you can access a common initial sequence of two structures in a union (ie, the destructor_ field) only if both structures are standard-layout types--at least according to a non-normative note in 9.5.1. According to 9.0.7, a standard-layout type can't have any non-static data members with non-standard layout. So if either A or B is not standard-layout, then it becomes illegal to access destructor_ in the wrong union.

A loophole would seem to be to make union_entry standard layout by turning value_ in to an alignas(T) char[sizeof(T)] . Nothing in 9.0.7 appears to rule out the use of the alignas . Thus, my question: Is the following a standard-layout type for any type T ? And hence can value_ to be cast to T& to emulate the previous example, while still allowing destructor_ to be used in a non-active union_entry ?

template<typename T> struct union_entry {
  void (*destructor_)(void *);
  alignas(T) char value_[sizeof(T)];
}

In both clang-3.8.1 and g++-6.2.1, std::is_standard_layout suggests union_entry<T> is standard layout even when T is not. Here's a complete working example of how I would like to use this technique:

#include <cassert>
#include <iostream>
#include <new>
#include <string>

using namespace std;

template<typename T> struct union_entry {
  void (*destructor_)(void *);
  alignas(T) char value_[sizeof(T)];

  union_entry() : destructor_(nullptr) {}
  ~union_entry() {}   // Just to cause error in unions w/o destructors

  void select() {
    if (destructor_)
      destructor_(this);
    destructor_ = destroy_helper;
    new (static_cast<void *>(value_)) T{};
  }
  T &get() {
    assert(destructor_ == destroy_helper);
    return *reinterpret_cast<T *>(value_);
  }

private:
  static void destroy_helper(void *_p) {
    union_entry *p = static_cast<union_entry *>(_p);
    p->get().~T();
    p->destructor_ = nullptr;
  }
};

union U {
  union_entry<int> i;
  union_entry<string> s;
  U() : i() {}
  ~U() { if (i.destructor_) i.destructor_(this); }
};

int
main()
{
  U u;
  u.i.select();
  u.i.get() = 5;
  cout << u.i.get() << endl;
  u.s.select();
  u.s.get() = "hello";
  cout << u.s.get() << endl;
  // Notice that the string in u.s is destroyed by calling
  // u.i.destructor_, not u.s.destructor_
}

Answer 1

Thanks to @Arvid, who pointed me to std::aligned_storage , I believe there is a definitive (albeit non-normative) answer in section 20.10.7.6 of the standard (which I assume is the same as N4141).

First, Table 57 says of aligned_storage "The member typedef type shall be a POD type ...", where 9.0.10 makes clear "A POD struct is a non-union class that is both a trivial class and a standard-layout class ."

Next, 20.10.7.6.1 gives a non-normative example implementation:

template <std::size_t Len, std::size_t Alignment>
struct aligned_storage {
  typedef struct {
    alignas(Alignment) unsigned char __data[Len];
  } type;
};

So clearly the use of alignas does not prevent a type from being standard-layout.