C++17 标准是否保证联合的地址与其成员的地址相同？
Does the C++17 standard guarantee that the address of a union is the same as the address of its members?
我目前正在编写池分配器的程序。我的问题归结为以下代码：
template <typename T>
union myUnion {
T data;
myUnion<T>* nextUnion;
};
void someFunction(){
myUnion<T> mu;
T* t = new (std::addressof(mu.data)) T();
//some code
myUnion<T>* mu2 = reinterpret_cast<myUnion<T>*>(t);
}
mu的地址总是和mu2一样吗？
是。
9.2/19（N4659 中为 12.2/24）：
If a standard-layout class object has any non-static data members, its address is the same as the address of its first non-static data member.
如果工会本身是标准布局，那么工会的地址与其成员的地址相同。
感谢9.5/1（N4659为12.3/2），会员地址都一样：
Each non-static data member is allocated as if it were the sole member of a
struct. All non-static data members of a union object have the same address.
仅仅询问两个对象是否具有相同的地址是不够的。例如，数组及其第一个元素保证具有相同的地址，but you cannot cast one to the other.
幸运的是，union 及其成员保证是 pointer-interconvertible，不管它们是否是标准布局。
我目前正在编写池分配器的程序。我的问题归结为以下代码：
template <typename T>
union myUnion {
T data;
myUnion<T>* nextUnion;
};
void someFunction(){
myUnion<T> mu;
T* t = new (std::addressof(mu.data)) T();
//some code
myUnion<T>* mu2 = reinterpret_cast<myUnion<T>*>(t);
}
mu的地址总是和mu2一样吗？
是。
9.2/19（N4659 中为 12.2/24）：
If a standard-layout class object has any non-static data members, its address is the same as the address of its first non-static data member.
如果工会本身是标准布局，那么工会的地址与其成员的地址相同。
感谢9.5/1（N4659为12.3/2），会员地址都一样：
Each non-static data member is allocated as if it were the sole member of a struct. All non-static data members of a union object have the same address.
仅仅询问两个对象是否具有相同的地址是不够的。例如，数组及其第一个元素保证具有相同的地址，but you cannot cast one to the other.
幸运的是，union 及其成员保证是 pointer-interconvertible，不管它们是否是标准布局。