在没有虚拟方法或超类的类中，是否可以安全地假设（第一个成员变量的地址）==这样？ [英] In a class with no virtual methods or superclass, is it safe to assume (address of first member variable) == this?

问题描述

我制作了一个私有API，该API假定该类中第一个成员对象的地址将与该类的this-pointer相同...这样，该成员对象就可以轻松地得出该对象的指针。

鉴于我愿意确保容器类不会从任何超类继承，所以它是其中的一员。没有任何虚拟方法，执行此技巧的成员对象将是第一个声明的成员对象，该假设对任何C ++编译器都有效，还是我需要使用offsetof（）运算符（或类似方法）保证正确性？

换句话说，下面的代码符合我在g ++下的预期，但是它可以在任何地方使用吗？

  class MyContainer 
 {
 public：
 MyContainer（）{} 
〜MyContainer（）{} //非-虚拟dtor 
 
私人：
类MyContained 
 {
公共：
 MyConta ined（）{} 
〜MyContained（）{} 
 
 //假设声明包含对象的唯一位置是m_contained 
 //（如下所示），预期在任何C ++编译器上？ 
 MyContainer * GetPointerToMyContainer（）
 {
 return reinterpret_cast< MyContainer *>（this）; 
} 
}; 
 
 MyContained m_contained; //必须是MyContainer中声明的第一个成员项目
 int m_foo; //其他成员项可以在m_contained 
 float m_bar之后声明； 
}; 
  

解决方案

似乎当前标准仅对POD类型保证了这一点

9.2.17

指向POD结构对象的指针，经过适当转换的
指向其
的初始成员（如果该成员是
位域，则指向其所在的单元
），反之亦然。 [注意：
因此，在POD结构对象
中可能存在未命名的
填充，但在其开始处没有，为必要的对齐
。 ]

但是，C ++ 0x标准似乎将这种保证扩展到了标准布局结构对象

标准布局类是
的类，

-不具有
类型的非标准布局类（或此类类型的
数组）或引用的静态数据成员

-没有虚函数（10.3）和
没有虚拟基类（10.1），

-对所有非静态对象具有相同的访问控制（子句
11）数据成员

-没有非标准布局的基础
类，

在派生度最高的类别中没有非静态数据
成员，并且在最多一个具有非静态
数据成员的基类中没有
，或者没有具有非静态数据
的基类静态数据成员，并且

-没有与第一个非静态数据成员相同类型的基类
。

A stan dard-layout结构是一个
标准布局类，定义了
class-key结构或class-key
类。

该假设很可能在实践中成立（前者不仅具有这些区别，尽管这可能是故意的）？

I made a private API that assumes that the address of the first member-object in the class will be the same as the class's this-pointer... that way the member-object can trivially derive a pointer to the object that it is a member of, without having to store a pointer explicitly.

Given that I am willing to make sure that the container class won't inherit from any superclass, won't have any virtual methods, and that the member-object that does this trick will be the first member object declared, will that assumption hold valid for any C++ compiler, or do I need to use the offsetof() operator (or similar) to guarantee correctness?

To put it another way, the code below does what I expect under g++, but will it work everywhere?

class MyContainer
{
public:
   MyContainer() {}
   ~MyContainer() {}  // non-virtual dtor

private:
   class MyContained
   {
   public:
      MyContained() {}
      ~MyContained() {}

      // Given that the only place Contained objects are declared is m_contained
      // (below), will this work as expected on any C++ compiler?
      MyContainer * GetPointerToMyContainer()
      {
         return reinterpret_cast<MyContainer *>(this);
      }
   };

   MyContained m_contained;  // MUST BE FIRST MEMBER ITEM DECLARED IN MyContainer
   int m_foo;                // other member items may be declared after m_contained
   float m_bar;
};

解决方案

It seems the current standard guarantees this only for POD types.

9.2.17

A pointer to a POD-struct object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa. [Note: There might therefore be unnamed padding within a POD-struct object, but not at its beginning, as necessary to achieve appropriate alignment. ]

However, the C++0x standard seems to extend this guarantee to "standard-layout struct object"

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,

— has no virtual functions (10.3) and no virtual base classes (10.1),

— has the same access control (Clause 11) for all non-static data members,

— has no non-standard-layout base classes,

— 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 no base classes with non-static data members, and

— has no base classes of the same type as the first non-static data member.

A standard-layout struct is a standard-layout class defined with the class-key struct or the class-key class.

It is probably likely that the assumption holds in practice (and the former didn't just have these distinctions, though this could have been the intention)?

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