c++ - 包含包装类型和类型本身的 union 是否有任何保证？

Question

我可以将一个 T 和一个包装的 T 放在一个 union 中并根据需要检查它们吗？

union Example {
    T value;
    struct Wrapped { 
       T wrapped;
    } wrapper;
};

// for simplicity T = int

Example ex;
ex.value = 12;
cout << ex.wrapper.wrapped; // ?

C++11 标准只保证公共(public)初始序列的保存检查，但 value 不是 struct。我猜想答案是否，因为wrapped types aren't even guaranteed to be memory compatible to their unwrapped counterpart和 accessing inactive members is only well-defined on common initial sequences .

Answer 1

我认为这是未定义的行为。

[class.mem]给我们:

The common initial sequence of two standard-layout struct types is the longest sequence of non-static data members and bit-fields in declaration order, starting with the first such entity in each of the structs, such that corresponding entities have layout-compatible types and either neither entity is a bit-field or both are bit-fields with the same width. [...]

In a standard-layout union with an active member of struct type T1, it is permitted to read a non-static data member m of another union member of struct type T2 provided m is part of the common initial sequence of T1 and T2; the behavior is as if the corresponding member of T1 were nominated.

如果 T 不是标准布局结构类型，这显然是未定义的行为。 (请注意，int 不是标准布局结构类型，因为它根本不是类类型)。

但即使对于标准布局结构类型，构成“公共(public)初始序列”的内容也严格基于非静态数据成员。即 T 和 struct { T val; 没有共同的初始序列 - 根本没有共同的数据成员!

因此，这里:

template <typename T>
union Example {
    T value;
    struct Wrapped { 
       T wrapped;
    } wrapper;
};


Example<int> ex;
ex.value = 12;
cout << ex.wrapper.wrapped; // (*)

您正在访问一个不活跃的联盟成员。这是未定义的。