如何声明相同类的成员向量? [英] How can I declare a member vector of the same class?

问题描述

到底为什么下面的代码起作用?

Why on earth does the following piece of code work?

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

A是不完整的类型，对吗?如果有一个A * s的向量，我会理解的.但是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

此纸张被导入C ++ 17 ，它允许在某些STL容器中使用不完整的类型.在此之前，它是未定义的行为.引用本文:

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*，以继续采用该方法–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.

关于标准的变化(重点是我的):

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

如果实例化vector，则可能会使用不完整的类型T. allocator 满足 allocator-completeness-requirements (17.6.3.5.1). T 必须在结果的任何成员之前完成 向量的专业化.

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.

因此，如果在实例化std::vector<T, Allocator>时将默认的std::allocator<T>保留在原位，则该文件将始终使用不完整的类型T起作用.否则，这取决于您的 Allocator 是否可以使用不完整的类型T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，对吗?如果有一个A * s的向量，我会理解的.但是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;
};

从技术上讲，除了size，capacity以及可能的allocator外，std::vector仅需要保留指向通过分配器管理的A动态数组的指针. (并且在编译时就知道了指针的大小.)

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，一个实现可能看起来像这样:

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;
    };

}

这篇关于如何声明相同类的成员向量?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！