我们需要一个可访问的复制构造函数用于C ++ 98/03中的值初始化？ [英] Do we need an accessible copy constructor for value initialization in C++98/03?

问题描述

此问题仅指前C ++ 11 。考虑下面看似破碎的代码：

  struct X 
 {
 X用户提供的构造函数
 private：
 X（const X&）{} 
}; 
 
 int main（）
 {
 X x = X（）; 
} 
  

Live on Coliru

根据 cppreference.com 在C ++ 11之前的默认ctor将被调用：

价值初始化的效果是：

1）如果T是至少有一个用户提供的类类型

p>这似乎暗示复制ctor不一定需要可访问。这是正确与否？

解决方案

上面的代码不能编译，因此看起来值初始化不需要它，但你需要一个可访问的副本构造函数来执行此操作：

  X x = X 
  

这是复制初始化，需要一个可访问的复制构造函数。即使它不会调用该复制构造函数，它仍然需要存在。

C ++ 17可能是第一个需要解除的版本。

This question refers only to pre C++11. Consider the following seemingly broken code:

struct X
{
    X(){} // default user-provided constructor
private:
    X(const X&){}
};

int main()
{
    X x = X();
}

Live on Coliru

According to cppreference.com in pre C++11 the default ctor will be called:

The effects of value initialization are:

1) if T is a class type with at least one user-provided constructor of any kind, the default constructor is called;

...

This seem to imply that the copy ctor doesn't necessarily need to be accessible. Is this correct or not? The code above does not compile, so it seems that a copy ctor must be accessible.

解决方案

Value initialization doesn't require it, but you need an accessible copy constructor to do this:

X x = X();

That's copy initialization, which requires an accessible copy constructor. Even though it will not call that copy constructor, it still needs to exist.

C++17 might be the first version where that need will be lifted.

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