c++ - 幻影类型是否与原始类型具有相同的对齐方式？

Question

考虑以下包含一些环境值的结构:

struct environment_values {
  uint16_t humidity;
  uint16_t temperature;
  uint16_t charging;
};

我想为这些具有幻像类型*的值添加一些附加信息，并同时使它们的类型不同:

template <typename T, typename P>
struct Tagged {
    T value;
};

// Actual implementation will contain some more features
struct Celsius{};
struct Power{};
struct Percent{};

struct Environment {
  Tagged<uint16_t,Percent> humidity;
  Tagged<uint16_t,Celsius> temperature;
  Tagged<uint16_t,Power>   charging;
};

Environment的内存布局和environment_values一样吗？这是否也适用于混合类型布局，例如:

struct foo {
    uint16_t value1;
    uint8_t  value2;
    uint64_t value3;
}

struct Foo {
    Tagged<uint16_t, Foo>  Value1;
    Tagged<uint8_t , Bar>  Value2;
    Tagged<uint64_t, Quux> Value3;
}

到目前为止，我尝试过的所有类型都有以下断言:

template <typename T, typename P = int>
constexpr void check() {
    static_assert(alignof(T) == alignof(Tagged<T,P>), "alignment differs");
    static_assert(sizeof(T)  == sizeof(Tagged<T,P>),  "size differs");
}

// check<uint16_t>(), check<uint32_t>(), check<char>() …

由于标记和未标记变体的大小也相同，我猜答案应该是肯定的，但我想有一些确定性。

^{* 我不知道在 C++ 中如何调用这些标记值。 “强类型类型定义”？我取自 Haskell 的名字。}

Answer 1

标准在 [basic.align]/1 中提及:

Object types have alignment requirements (3.9.1, 3.9.2) which place restrictions on the addresses at which an object of that type may be allocated. An alignment is an implementation-defined integer value representing the number of bytes between successive addresses at which a given object can be allocated. An object type imposes an alignment requirement on every object of that type; stricter alignment can be requested using the alignment specifier (7.6.2).

此外，[basic.compound]/3 ，提及:

The value representation of pointer types is implementation-defined. Pointers to layout-compatible types shall have the same value representation and alignment requirements (6.11). [Note: Pointers to over-aligned types (6.11) have no special representation, but their range of valid values is restricted by the extended alignment requirement].

因此，可以保证布局兼容的类型具有相同的对齐方式。

struct { T m; } 和 T 布局不兼容。

正如所指出的 here ，为了使两个元素布局兼容，那么它们都必须是标准布局类型，并且它们的非静态数据成员必须以相同的类型和相同的顺序出现。

struct { T m; } 只包含一个 T，但 T 是一个 T 所以它不能包含一个 T 作为它的第一个非静态数据成员。