如何声明同一个类的成员向量？

Question

Why on earth does the following piece of code work?

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

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.

Answer 1

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:

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.

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

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.

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 .

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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.

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

class A{
    A* subAs;
};

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.)

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;
    };

}