确定模板中结构或元组的成员偏移量

我想编写一个将表写入HDF5文件的模板函数。 签名应类似

template<typename record> void writeTable(const std::vector<record>& data);

其中Record是结构，或

template<typename... elements> 
    void writeTable(const std::vector<std::tuple<elements...>>& data);

实际实现将有更多参数来确定目标等。

要编写数据，我需要定义一个HDF5复合类型，它包含成员的名称和偏移量。通常您会使用HOFFSET宏来获取字段偏移量，但由于我事先不知道结构字段，所以我不能这样做。

到目前为止，我尝试的是从typeName包构造一个结构类型。朴素的实现没有标准的布局，但实现here有。剩下的就是获得成员的偏移量。我想将参数包展开为具有偏移量的初始化列表：

#include <vector>

template<typename... members> struct record {};

template<typename member, typename... members> struct record<member, members...> : 
    record<members...> {
  record(member m, members... ms) : record<members...>(ms...), tail(m) {}
  member tail;
};

template<typename... Args> void 
    make_table(const std::string& name, const std::vector<record<Args...>>& data) {
  using record_type = record<Args...>;
  std::vector<size_t> offsets = { get_offset(record_type,Args)... };
}

int main() {
  std::vector<record<int, float>> table = { {1, 1.0}, {2, 2.0} };
  make_table("table", table);
}

get_offset是否有可能的实现？我不这么认为，因为在record<int, int>的情况下，它是模棱两可的。还有别的办法吗？

或者有没有其他方法可以解决这个问题？

---

解决方案

计算偏移量非常简单。给定类型为T0、T1的元组...TN.T0的偏移量是0(只要您在char数组上使用alignas(T0))。T1的偏移量是向上舍入到alignof(T1)的sizeof(T0)。

一般TB(位于TA之后)的偏移量为round_up(offset_of<TA>() + sizeof(TA), alignof(TB))。

计算std::tuple中元素的偏移量可以如下所示：

constexpr size_t roundup(size_t num, size_t multiple) {
  const size_t mod = num % multiple;
  return mod == 0 ? num : num + multiple - mod;
}

template <size_t I, typename Tuple>
struct offset_of {
  static constexpr size_t value = roundup(
    offset_of<I - 1, Tuple>::value + sizeof(std::tuple_element_t<I - 1, Tuple>),
    alignof(std::tuple_element_t<I, Tuple>)
  );
};

template <typename Tuple>
struct offset_of<0, Tuple> {
  static constexpr size_t value = 0;
};

template <size_t I, typename Tuple>
constexpr size_t offset_of_v = offset_of<I, Tuple>::value;

这是一个测试套件。正如您从第一个测试中看到的，元素的对齐被考虑在内。

static_assert(offset_of_v<1, std::tuple<char, long double>> == 16);
static_assert(offset_of_v<2, std::tuple<char, char, long double>> == 16);
static_assert(offset_of_v<3, std::tuple<char, char, char, long double>> == 16);
static_assert(offset_of_v<4, std::tuple<char, char, char, char, long double>> == 16);

static_assert(offset_of_v<0, std::tuple<int, double, int, char, short, long double>> == 0);
static_assert(offset_of_v<1, std::tuple<int, double, int, char, short, long double>> == 8);
static_assert(offset_of_v<2, std::tuple<int, double, int, char, short, long double>> == 16);
static_assert(offset_of_v<3, std::tuple<int, double, int, char, short, long double>> == 20);
static_assert(offset_of_v<4, std::tuple<int, double, int, char, short, long double>> == 22);
static_assert(offset_of_v<5, std::tuple<int, double, int, char, short, long double>> == 32);

我在上面的测试中硬编码了偏移量。如果以下测试成功，则偏移量正确。

static_assert(sizeof(char) == 1 && alignof(char) == 1);
static_assert(sizeof(short) == 2 && alignof(short) == 2);
static_assert(sizeof(int) == 4 && alignof(int) == 4);
static_assert(sizeof(double) == 8 && alignof(double) == 8);
static_assert(sizeof(long double) == 16 && alignof(long double) == 16);

std::tuple似乎按顺序存储它的元素(没有对它们进行排序以优化填充)。以下测试证明了这一点。我不认为该标准要求以这种方式实现std::tuple，因此我不认为以下测试一定会成功。

template <size_t I, typename Tuple>
size_t real_offset(const Tuple &tup) {
  const char *base = reinterpret_cast<const char *>(&tup);
  return reinterpret_cast<const char *>(&std::get<I>(tup)) - base;
}

int main(int argc, char **argv) {
  using Tuple = std::tuple<int, double, int, char, short, long double>;
  Tuple tup;
  assert((offset_of_v<0, Tuple> == real_offset<0>(tup)));
  assert((offset_of_v<1, Tuple> == real_offset<1>(tup)));
  assert((offset_of_v<2, Tuple> == real_offset<2>(tup)));
  assert((offset_of_v<3, Tuple> == real_offset<3>(tup)));
  assert((offset_of_v<4, Tuple> == real_offset<4>(tup)));
  assert((offset_of_v<5, Tuple> == real_offset<5>(tup)));
}

既然我已经完成了所有这些工作，real_offset函数是否适合您的需求？

---

这是访问char[]和offset_of的元组的最小实现。但由于reinterpret_cast，这是未定义的行为。即使我在相同的字节中构造对象并在相同的字节中访问对象，它仍然是UB。有关所有标准代码，请参阅this answer。它可以在您能找到的所有编译器上运行，但它是UB，所以无论如何都要使用它。这个元组是标准布局(与std::tuple不同)。如果元组的元素都是普通可复制的，则可以删除复制和移动构造函数，并将其替换为memcpy。

template <typename... Elems>
class tuple;

template <size_t I, typename Tuple>
struct tuple_element;

template <size_t I, typename... Elems>
struct tuple_element<I, tuple<Elems...>> {
  using type = std::tuple_element_t<I, std::tuple<Elems...>>;
};

template <size_t I, typename Tuple>
using tuple_element_t = typename tuple_element<I, Tuple>::type;

template <typename Tuple>
struct tuple_size;

template <typename... Elems>
struct tuple_size<tuple<Elems...>> {
  static constexpr size_t value = sizeof...(Elems);
};

template <typename Tuple>
constexpr size_t tuple_size_v = tuple_size<Tuple>::value;

constexpr size_t roundup(size_t num, size_t multiple) {
  const size_t mod = num % multiple;
  return mod == 0 ? num : num + multiple - mod;
}

template <size_t I, typename Tuple>
struct offset_of {
  static constexpr size_t value = roundup(
    offset_of<I - 1, Tuple>::value + sizeof(tuple_element_t<I - 1, Tuple>),
    alignof(tuple_element_t<I, Tuple>)
  );
};

template <typename Tuple>
struct offset_of<0, Tuple> {
  static constexpr size_t value = 0;
};

template <size_t I, typename Tuple>
constexpr size_t offset_of_v = offset_of<I, Tuple>::value;

template <size_t I, typename Tuple>
auto &get(Tuple &tuple) noexcept {
  return *reinterpret_cast<tuple_element_t<I, Tuple> *>(tuple.template addr<I>());
}

template <size_t I, typename Tuple>
const auto &get(const Tuple &tuple) noexcept {
  return *reinterpret_cast<tuple_element_t<I, Tuple> *>(tuple.template addr<I>());
}

template <typename... Elems>
class tuple {
  alignas(tuple_element_t<0, tuple>) char storage[offset_of_v<sizeof...(Elems), tuple<Elems..., char>>];
  using idx_seq = std::make_index_sequence<sizeof...(Elems)>;

  template <size_t I>
  void *addr() {
    return static_cast<void *>(&storage + offset_of_v<I, tuple>);
  }

  template <size_t I, typename Tuple>
  friend auto &get(const Tuple &) noexcept;

  template <size_t I, typename Tuple>
  friend const auto &get(Tuple &) noexcept;

  template <size_t... I>
  void default_construct(std::index_sequence<I...>) {
    (new (addr<I>()) Elems{}, ...);
  }
  template <size_t... I>
  void destroy(std::index_sequence<I...>) {
    (get<I>(*this).~Elems(), ...);
  }
  template <size_t... I>
  void move_construct(tuple &&other, std::index_sequence<I...>) {
    (new (addr<I>()) Elems{std::move(get<I>(other))}, ...);
  }
  template <size_t... I>
  void copy_construct(const tuple &other, std::index_sequence<I...>) {
    (new (addr<I>()) Elems{get<I>(other)}, ...);
  }
  template <size_t... I>
  void move_assign(tuple &&other, std::index_sequence<I...>) {
    (static_cast<void>(get<I>(*this) = std::move(get<I>(other))), ...);
  }
  template <size_t... I>
  void copy_assign(const tuple &other, std::index_sequence<I...>) {
    (static_cast<void>(get<I>(*this) = get<I>(other)), ...);
  }

public:
  tuple() noexcept((std::is_nothrow_default_constructible_v<Elems> && ...)) {
    default_construct(idx_seq{});
  }
  ~tuple() {
    destroy(idx_seq{});
  }
  tuple(tuple &&other) noexcept((std::is_nothrow_move_constructible_v<Elems> && ...)) {
    move_construct(other, idx_seq{});
  }
  tuple(const tuple &other) noexcept((std::is_nothrow_copy_constructible_v<Elems> && ...)) {
    copy_construct(other, idx_seq{});
  }
  tuple &operator=(tuple &&other) noexcept((std::is_nothrow_move_assignable_v<Elems> && ...)) {
    move_assign(other, idx_seq{});
    return *this;
  }
  tuple &operator=(const tuple &other) noexcept((std::is_nothrow_copy_assignable_v<Elems> && ...)) {
    copy_assign(other, idx_seq{});
    return *this;
  }
};

---

或者，您可以使用此函数：

template <size_t I, typename Tuple>
size_t member_offset() {
  return reinterpret_cast<size_t>(&std::get<I>(*static_cast<Tuple *>(nullptr)));
}

template <typename Member, typename Class>
size_t member_offset(Member (Class::*ptr)) {
  return reinterpret_cast<size_t>(&(static_cast<Class *>(nullptr)->*ptr));
}

template <auto MemPtr>
size_t member_offset() {
  return member_offset(MemPtr);
}

再说一次，这是未定义的行为(因为nullptr取消引用和reinterpret_cast)，但它将在每个主要编译器中按预期工作。该函数不能为constexpr(即使成员偏移量是编译时计算)。