"friending" CRTP 继承中的基 class 是否也会影响子 class？

Question

在 attempt to answer another question 中，我想出了一个方案来强制 CRTP 基 class 的子级接受特定类型作为其构造函数中的参数：使参数类型的构造函数 private，将 CRTP 基 class 分配为 friend，并将参数类型声明为基 class 构造函数的参数。

然而，当我试图证明该方案通过访问冲突提供了所需的保护时，我发现即使参数类型的构造函数是私有的，子 class 也能够构造它：

template <typename T>
class SingletonBase {
  protected: class P { friend class SingletonBase<T>; P() = default; };
  public:
     SingletonBase(P) {} 
};

class Logger: public SingletonBase<Logger> {
  using BASE = SingletonBase<Logger>;
  public:
    Logger() : BASE{P{}} {} // WHY NO ACCESS VIOLATION?
};

这个 compiles without error，尽管我预计会发生访问冲突。为什么？

Answer 1

你所做的与你的friend声明无关！

如果删除 friend 代码编译也很好！

这是因为空 class 的默认构造函数是 public:

来自 C++11 标准：

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

如果你没有像这样的默认构造函数:

template <typename T>
class SingletonBase
{
    protected: 
        class P
        { 
            friend class SingletonBase<T>;
            P(int){ }
        };

    public:
        SingletonBase(P) {}
};

class Logger: public SingletonBase<Logger>
{
    using BASE = SingletonBase<Logger>;

    public:
    Logger() : BASE(P{1}) {} // WHY NO ACCESS VIOLATION?
};

您将遇到 "access" 违规，您会发现您的 friend 无效！：

main.cpp: In constructor 'Logger::Logger()':
main.cpp:10:17: error: 'SingletonBase<T>::P::P(int) [with T = Logger]' is private
                 P(int){ }
                 ^
main.cpp:22:28: error: within this context
         Logger() : BASE(P{1}) {} // WHY NO ACCESS VIOLATION?

Answer 2

Does “friending” the base class in CRTP inheritance affect the child class as well?

不，当然不是。友情不是世代相传的。为了说明这个问题，

首先，P::P()是一个默认的默认构造函数，它是一个trivial default constructor。

其次，P{} 是 value initialization (C++11 起)，

（强调我的）

2) if T is a class type with a default constructor that is neither user-provided nor deleted (that is, it may be a class with an implicitly-defined or defaulted default constructor), the object is zero-initialized and then it is default-initialized if it has a non-trivial default constructor;

请注意，只有 zero initialized here, not default initializated。 P 的私有默认构造函数根本不会被调用。

If T is an non-union class type, all base classes and non-static data members are zero-initialized, and all padding is initialized to zero bits. The constructors, if any, are ignored.

如果您将其显式更改为默认初始化，则会出现访问冲突错误。

Logger() : BASE{P()} {} // error: calling a private constructor of class 'SingletonBase<Logger>::P
//               ~~

简化演示

class X { X() = default; };

int main()
{
    X x1{}; // fine
    X x2;   // error: calling a private constructor of class 'X'
}

LIVE

解决方案

您可以提供一个用户定义的默认构造函数，这是一个非平凡的构造函数，以更改值初始化的行为。

template <typename T>
class SingletonBase {
  protected: 
    class P { 
      friend class SingletonBase<T>; 
      P() {} // user-defined default constructor
    };
  public:
    SingletonBase(P) {} 
};

class Logger: public SingletonBase<Logger> {
  using BASE = SingletonBase<Logger>;
  public:
    Logger() : BASE{P{}} {} // error: calling a private constructor of class 'SingletonBase<Logger>::P'
};

"friending" CRTP 继承中的基 class 是否也会影响子 class？

Does "friending" the base class in CRTP inheritance affect the child class as well?

c++

inheritance

private

friend

crtp