如何是一类的内存布局与结构体 [英] How is the memory layout of a class vs. a struct

问题描述

我来自C编程其中，在结构中的数据与前第一可变布局，然后是第二，第三等..

我现在的编程C ++和我使用的是类代替。我基本上要达到相同的，但我也想get / set方法，也可能是其它方法（我也想尝试做一个C ++风格和马爷学习新的东西）。

有保证例如公众变量将先在内存中，然后私有变量？

解决方案

  

有保证例如公众变量将先在
  内存则私有变量？

没有，这样的保证是的不可以制作 - C ++ 11标准，[class.mem] / 14：

  

是相同的访问（非工会）类的非静态数据成员
  控制（第11）分配以至于后来的成员有较高
  一类对象中的地址。 非静态分配的顺序
  具有不同的访问控制数据成员是不确定（11）。

所以

 一个结构
{
    INT I，J;
    性病::字符串str;私人的：    浮F;保护：    双D;
};
 

这只能保证，对于类型的给定对象 A ，

• I 比更小的地址Ĵ和


• Ĵ比更小的地址 STR

需要注意的是类键结构和类有没有关于差别布局凡：他们唯一的区别是访问-rights只存在在编译时。

---

  

这只是说的顺序，但不是第一个变量真正开始
  在首地址？让我们假设一个类继承没有

是的，但仅限于标准布局类。还有就是要求类必须满足以下是一个标准布局类，其中之一是，所有成员都具有相同的访问控制的行。结果
引用C ++ 14（同样适用于C ++ 11，但措辞更为间接的），[class.mem] / 19：

  

如果一个标准的布局类对象具有任何非静态数据成员，其
  地址是相同的其第一非静态数据的地址
  会员。否则，它的地址是相同的它的第一个的地址
  基classsubobject（如有的话）。 [注意的：有可能因此受到
  根据需要一个标准布局结构对象内无名填充，但不是在它的开头，以实现适当的对准。 - 注完的]

[类] / 7：

  

一个的标准布局类的是一类：

  
  

  
• 有型非标准布局类的没有非静态数据成员（或这些类型的数组）或引用，

  
• 没有虚函数（10.3），没有虚基类（10.1），

  
• 有相同的访问控制（第11），所有的非静态数据成员，

  
• 没有非标准布局基类，

  
• 或者具有在最派生类，并在与非静态数据成员最多有一个基类没有非静态数据成员，或者没有基
    类与非静态数据成员，以及

  
• 具有相同的类型与第一非静态数据成员的无基类。 ¹¹⁰

  

  
  

^{110）这确保两个子对象具有相同的类型，而且属于同一最派生的对象是不
  在同一个地址（5.10）。}

分配

I come from C programming where the data in a struct is laid out with the top variable first, then the second, third and so on..

I am now programming in C++ and I am using a class instead. I basically want to achieve the same, but I also want get/set methods and also maybe other methods (I also want to try do it in a C++ style and maye learn something new).

Is there a guarantee e.g. that the public variables will be first in memory then the private variable?

解决方案

Is there a guarantee e.g. that the public variables will be first in memory then the private variable?

No, such a guarantee is not made - C++11 standard, [class.mem]/14:

Nonstatic data members of a (non-union) class with the same access control (Clause 11) are allocated so that later members have higher addresses within a class object. The order of allocation of non-static data members with different access control is unspecified (11).

So

struct A
{
    int i, j;
    std::string str;

private:

    float f;

protected:

    double d;
};

It is only guaranteed that, for a given object of type A,

• i has a smaller address than j and
• j has a smaller address than str

Note that the class-keys struct and class have no difference regarding layout whatsoever: Their only difference are access-rights which only exist at compile-time.

---

It only says the order, but not that the first variable actually start at the "first address"? Lets assume a class without inheritance.

Yes, but only for standard-layout classes. There is a row of requirements a class must satisfy to be a standard-layout class, one of them being that all members have the same access-control.
Quoting C++14 (the same applies for C++11, but the wording is more indirect), [class.mem]/19:

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. Otherwise, its address is the same as the address of its first base classsubobject (if any). [ Note: There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning, as necessary to achieve appropriate alignment. — end note ]

[class]/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,
• 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. ¹¹⁰

^{110) This ensures that two subobjects that have the same class type and that belong to the same most derived object are not allocated at the same address (5.10).}

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