c++ - 为什么带有引用的类不遵守标准布局？

Question

执行以下代码:

#include <iostream>
#include <type_traits>

struct s_ref {
    int &foo;
};

struct s_ptr {
    int *foo;
};

int main(int argc, char *argv[])
{
    std::cout << "s_ref is_standard_layout:" << std::is_standard_layout<struct s_ref>::value << std::endl;
    std::cout << "s_ptr is_standard_layout:" << std::is_standard_layout<struct s_ptr>::value << std::endl;
    return 0;
}

结果:

s_ref is_standard_layout:0
s_ptr is_standard_layout:1

基于标准布局的使用(即:“标准布局类型对于与用其他编程语言编写的代码进行通信很有用”)这是有道理的，但我不确定违反了哪个规则:

A standard-layout class is a class (defined with class, struct orunion) that:

• has no virtual functions and no virtual base classes.

• has the same access control (private, protected, public) for all its non-static data members.

• either has no non-static data members in the most derived class and at most one base class with non-static data members, or has nobase classes with non-static data members.

• its base class (if any) is itself also a standard-layout class.

• And, has no base classes of the same type as its first non-static datamember.

编辑:引用来自:http://www.cplusplus.com/reference/type_traits/is_standard_layout/ , 但是 http://en.cppreference.com/w/cpp/concept/StandardLayoutType也是类似的。

Answer 1

C++ 标准的标准布局类概念的要点是，此类类的实例可以作为字节可靠地访问或复制到字节，正如 C++11 标准所指出的那样在其第 9/9 节中，创建了这样一个类

” useful for communicating with code written in other programming languages

但是，C++ 标准根本不需要引用来使用存储。它不是一个对象。你不能拿它的地址。所以它不能(可靠地)复制到字节，或作为字节访问。因此它与标准布局类的概念不兼容。

在正式场合，

C++11 §9/7:

” A standard-layout class is a class that:
— has no non-static data members of type non-standard-layout class (or array of such types) or reference,