c++ - 在常量表达式中访问 union 成员时出错

Question

当我遇到问题时，我正在用 union 做一些实验。

union U
{
  // struct flag for reverse-initialization of each byte
  struct rinit_t { };
  constexpr static const rinit_t rinit{};

  uint32_t dword;
  uint8_t byte[4];

  constexpr U() noexcept : dword{} { }

  constexpr U(uint32_t x) noexcept : dword{x} { }

  constexpr U(uint32_t x, const rinit_t&) noexcept : dword{}
  {
    U temp{x};
    byte[0] = temp.byte[3];
    byte[1] = temp.byte[2];
    byte[2] = temp.byte[1];
    byte[3] = temp.byte[0];
  }
};

这是我的示例实例:

constexpr U x{0x12345678, U::rinit};

我在带有 -std=c++14 、 -std=c++17 和 -std=c++2a 的 g++ 5.1 和 8.1 版本中遇到此错误:

accessing 'U::byte' member instead of initialized 'U::dword' member in constant expression

访问和分配成员 byte 的元素，无论是来自 temp 还是 this ，都会重现错误。即使 byte 和 byte 共享相同的地址，似乎 dword 也被编译器识别为“未初始化”成员。
我曾经修改过第二个构造函数:

constexpr U(uint32_t x) noexcept :
  byte{uint8_t(x), uint8_t(x >> 8), uint8_t(x >> 16), uint8_t(x >> 24)}
{ }

但是我在最终产生似乎是编译器错误后恢复了:

main.cpp:73:37: internal compiler error: in complete_ctor_at_level_p, at expr.c:5844
   constexpr U x{0x12345678, U::rinit};
                                     ^

Please submit a full bug report,
with preprocessed source if appropriate.
See <http://tdm-gcc.tdragon.net/bugs> for instructions.

对于我当前的修复，我添加了一个转换器:

// converts the value of a uint32_t to big endian format
constexpr static uint32_t uint32_to_be(uint32_t x)
{
  return ( (x >> 24) & 0xFF)       |
         ( (x << 8)  & 0xFF0000)   |
         ( (x >> 8)  & 0xFF00)     |
         ( (x << 24) & 0xFF000000);
}

我修改了第三个构造函数:

constexpr U(uint32_t x, const rinit_t&) noexcept : dword{uint32_to_be(x)} { }

我只是好奇为什么我会收到错误。谁能帮我理解这个问题？
更新:
根据我最近的测试，在 constexpr union 构造函数中，我不能使用不在初始化列表中的非静态数据成员。因此，我添加了一些 struct 标志来显式指定某个非静态数据成员的初始化。

// struct flag to explicitly specify initialization of U::dword
struct init_dword_t { };
constexpr static const init_dword_t init_dword{};

// struct flag to explicitly specify initialization of U::byte
struct init_byte_t { };
constexpr static const init_byte_t init_byte{};

有了它，我还为这种初始化添加了新的构造函数。这里有些例子:

constexpr U(const init_byte_t&) noexcept : byte{} { }

// for some reason, this version does not reproduce the internal compiler error
constexpr U(uint32_t x, const init_byte_t&) noexcept :
  byte{uint8_t(x), uint8_t(x >> 8), uint8_t(x >> 16), uint8_t(x >> 24)}
{ }

constexpr U(uint32_t x, const init_byte_t&, const rinit_t&) noexcept :
  byte{uint8_t(x >> 24), uint8_t(x >> 16), uint8_t(x >> 8), uint8_t(x)}
{ }

如果有人可以为此提供更好的解决方案，那就太好了。

Answer 1

您的 constexpr函数在c++20之前无效，因为它违反了以下rule :

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

an assignment expression or invocation of an assignment operator ([class.copy]) that would change the active member of a union;

从 c++20 开始，此限制已得到澄清 here :

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

an invocation of an implicitly-defined copy/move constructor or copy/move assignment operator for a union whose active member (if any) is mutable, unless the lifetime of the union object began within the evaluation of E;

我相信这使您的代码有效。