[英]std::vector, std::move and pointer invalidation
我的问题与以下模式有关...我想使用以下模式来构造一个昂贵的构造SomeData
,然后将其移至UsesData
。
所以问题是...
是否保证ud.dat.m_ptrs
的指针仍然有效?
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 } };
//...
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.