如何声明同一类的成员向量?

How can I declare a member vector of the same class?(如何声明同一类的成员向量?)

问题描述

到底为什么下面这段代码有效?

Why on earth does the following piece of code work?

struct A {
    std::vector<A> subAs;
};

A 是不完整的类型，对吧?如果有 A*s 的向量，我会理解.但在这里我不明白它是如何工作的.这似乎是一个递归定义.

A is an incomplete type, right? If there was a vector of A*s I would understand. But here I don't understand how it works. It seems to be a recursive definition.

推荐答案

这个 paper 被引入 C++17 允许在某些 STL 容器中使用不完整的类型.在此之前，它是未定义的行为.引用论文:

This paper was adopted into C++17 which allows incomplete types to be used in certain STL containers. Prior to that, it was Undefined Behavior. To quote from the paper:

基于 Issaquah 会议的讨论，我们实现了共识继续*采用该方法——不完整的容器类型"，但将范围限制为 std::vector、std::list 和std::forward_list，作为第一步.

Based on the discussion on the Issaquah meeting, we achieved the consensus to proceed* with the approach – "Containers of Incomplete Types", but limit the scope to std::vector, std::list, and std::forward_list, as the first step.

至于标准的变化(重点是我的):

And as for the changes in the standard (emphasis mine):

一个不完整的类型 T 可以在实例化 vector 时使用，如果allocator 满足allocator-completeness-requirements(17.6.3.5.1).T 应在产生的任何成员之前完成引用了向量的特化.

An incomplete type T may be used when instantiating vector if the allocator satisfies the allocator-completeness-requirements (17.6.3.5.1). T shall be complete before any member of the resulting specialization of vector is referenced.

所以，如果你在实例化 std::vector 时保留默认的 std::allocator，那么根据论文，它将始终使用不完整的类型 T ；否则，这取决于您的 Allocator 是否可以使用不完整的类型 T 进行实例化.

So, there you have it, if you leave the default std::allocator<T> in place when instantiating the std::vector<T, Allocator>, then it will always work with an incomplete type T according to the paper; otherwise, it depends on your Allocator being instantiable with an incomplete type T.

A 是不完整的类型，对吧?如果有 A*s 的向量，我会理解.但在这里我不明白它是如何工作的.这似乎是一个递归定义.

A is an incomplete type, right? If there was a vector of A*s I would understand. But here I don't understand how it works. It seems to be a recursive definition.

那里没有递归.以极其简化的形式，它类似于:

There is no recursion there. In an extremely simplified form, it's similar to:

class A{
    A* subAs;
};

技术上，除了size、capacity和可能的allocator，std::vector只需要保持一个指向 A 动态数组的指针，它通过它的分配器管理.(并且指针的大小在编译时是已知的.)

Technically, apart from size, capacity and possibly allocator, std::vector only needs to hold a pointer to a dynamic array of A it manages via its allocator. (And the size of a pointer is known at compile time.)

因此，实现可能如下所示:

So, an implementation may look like this:

namespace std{

    template<typename T, typename Allocator = std::allocator<T>>
    class vector{

        ....

        std::size_t m_capacity;
        std::size_t m_size;
        Allocator m_allocator;
        T* m_data;
    };

}

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