c++ - public/private/protected 是否会改变结构在内存中的排列？

Question

编辑:我刚刚意识到提出这个问题的一种更简单的方法:

Given the following two structs:

class Thing {public: int a; public: int b; public: int c;}

class Thing {public: int a, private: int b; public: int c;}

Are the members a, b, and c guaranteed to be in the same order in memory for both of these definitions?

---

老问题

假设我们在 fileA.cpp 中有这段 C++ 代码:

class Thing
{
public:
    int a;
    double num;

    Thing()
    {
        b = 10;
    }

    float getB()
    {
        return b;
    }

private:
    float b;
    Thing * other;
}

void doSomething(Thing thing);

int main()
{
    Thing thing;
    doSomething(thing);
    std::cout << thing.b;
}

假设我们在 fileB.cpp 中有这段代码:

class Thing
{
public:
    int a;
    double num;

    Thing()
    {
        b = 10;
    }

    float getB()
    {
        return b;
    }

    float b;

private:
    Thing * other;
}

void doSomething(Thing thing)
{
    thing.b = 30;
}

假设编译器不会报错，这段代码会按预期工作吗？也就是说，结构数据的排列是否独立于某些组件是公共(public)的、私有(private)的还是 protected ？

编辑:为了使它更明显，Thing 的两个定义之间的唯一区别是 float b; 是私有(private)的在 fileA.cpp 中，但在 fileB.cpp 中公开。

Answer 1

标准没有做出这样的保证。您只有标准布局类的布局保证:

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.

(C++14，[类]¶7)

如果一个类是标准布局，则其布局定义良好(两个具有布局兼容初始序列的标准布局类可以通过 union 读取彼此的布局兼容成员)。

但是，这里不是这种情况，因为您在整个类中有不同的访问说明符。特别是，它明确指出

The order of allocation of non-static data members with different access control is unspecified

(C++14, [class.mem] ¶13)

---

话虽这么说，但我从未使用过任何利用标准提供的这种灵 active 的现实世界编译器 - 我知道的每个编译器都使用访问说明符进行编译时检查，但完全忽略它们就成员布局而言。