std::vector、std::move 和指针失效

Question

我的问题与以下模式有关... 我想用下面的模式构造一个昂贵的构造SomeData，然后将它移动到UsesData.

所以问题是...

ud.dat.m_ptrs中的指针是否保证仍然有效？

struct BigObject{};
struct SomeData
{
    SomeData() = default;

    SomeData(const SomeData &) = delete;
    SomeData & operator = (const SomeData &) = delete;

    SomeData(SomeData &&) = default;
    SomeData & operator = (SomeData &&) = default;

    std::vector<BigObject> m_data1; // big vector
    std::vector<BigObject> m_data2; // big vector

    // (m_ptrs.size() == m_data1.size() + m_data2.size())
    // points to elements in m_data1 and m_data2...
    std::vector<const BigObject * const> m_ptrs;
};

struct Builder
{
    Builder() = delete;
    Builder(const Builder &) = delete;
    Builder & operator=(const Builder &) = delete;
    Builder(Builder &&) = delete;
    Builder & operator=(Builder &&) = delete;

    Builder(int a)  
    {
        // makes sure BigObject vectors in SomeDate are constructed correctly
        // builds m_ptrs... vector of ptrs to m_data1 and m_data2
    }

    SomeData dat;
};

struct UsesData
{
    UsesData() = delete;
    UsesData(const UsesData &) = delete;
    UsesData & operator=(const UsesData &) = delete;
    UsesData(UsesData &&) = delete;
    UsesData & operator=(UsesData &&) = delete;

    UsesData(Builder && from) : dat{ std::move(from.dat) }
    {}

    const SomeData dat;
};

int main()
{
    UsesData ud{ Builder{ 1 } };
    //...
}

Answer 1

指针将保持有效。根据std::vector的move constructor的行为：

After container move construction (overload (6)), references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in *this. The current standard makes this guarantee via the blanket statement in §23.2.1[container.requirements.general]/12, and a more direct guarantee is under consideration via LWG 2321.

这意味着，在被移动之后，指针仍然有效并指向被移动到新std::vector的元素。

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