問題描述

到底為什么下面這段代碼有效?

Why on earth does the following piece of code work?

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

A 是不完整的類型，對吧?如果有 A*s 的向量，我會理解.但在這里我不明白它是如何工作的.這似乎是一個遞歸定義.

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.

推薦答案

這個 paper 被引入 C++17 允許在某些 STL 容器中使用不完整的類型.在此之前，它是未定義的行為.引用論文:

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 會議的討論，我們實現了共識繼續*采用該方法——不完整的容器類型"，但將范圍限制為 std::vector、std::list 和std::forward_list，作為第一步.

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.

至于標準的變化(重點是我的):

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

一個不完整的類型 T 可以在實例化 vector 時使用，如果allocator 滿足allocator-completeness-requirements(17.6.3.5.1).T 應在產生的任何成員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，如果你在實例化 std::vector 時保留默認的 std::allocator，那么根據論文，它將始終使用不完整的類型 T ；否則，這取決于您的 Allocator 是否可以使用不完整的類型 T 進行實例化.

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 是不完整的類型，對吧?如果有 A*s 的向量，我會理解.但在這里我不明白它是如何工作的.這似乎是一個遞歸定義.

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.

那里沒有遞歸.以極其簡化的形式，它類似于:

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

class A{
    A* subAs;
};

技術上，除了size、capacity和可能的allocator，std::vector只需要保持一個指向 A 動態數組的指針，它通過它的分配器管理.(并且指針的大小在編譯時是已知的.)

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，實現可能如下所示:

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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