为什么 deque::erase() 调用赋值运算符？

Question

正如标题所说，为什么deque在erase()时调用了包含类型的赋值运算符？我能理解为什么 vector 可能是因为 vector 中的元素在连续的内存中，但是由于 deque 不能保证连续的内存，为什么它会在删除它的一些元素时尝试移动它的元素。

此代码无法编译，因为我的 Contained 类型的赋值运算符被删除并且它没有移动构造函数。

#include <deque>

class Contained
{
public:
    Contained() = default;
    ~Contained() { }
    Contained(const Contained&) = delete;
    Contained& operator=(const Contained&) = delete;
};

class Container
{
public:
    Container()
    {
        for(int i = 0; i < 5; i++) { m_containerDS.emplace_back(); }
    }

    ~Container() { }

    void clear() 
    { 
        m_containerDS.erase(m_containerDS.begin(), m_containerDS.end()); 
    }

private:
    std::deque<Contained> m_containerDS;
};


int main()
{
    return 0;
}

MSVC 编译器发出此错误消息：

C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\xutility(2527): error C2280: 'Contained &Contained::operator =(const Contained &)' : attempting to reference a deleted function
1>          ConsoleApplication13.cpp(12) : see declaration of 'Contained::operator ='
1>          C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\xutility(2548) : see reference to function template instantiation '_BidIt2 std::_Move_backward<_BidIt1,_BidIt2>(_BidIt1,_BidIt1,_BidIt2,std::_Nonscalar_ptr_iterator_tag)' being compiled
1>          with
1>          [
1>              _BidIt2=std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>
1>  ,            _BidIt1=std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\deque(1622) : see reference to function template instantiation '_BidIt2 std::_Move_backward<std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>,std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>>(_BidIt1,_BidIt1,_BidIt2)' being compiled
1>          with
1>          [
1>              _BidIt2=std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>
1>  ,            _BidIt1=std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>>
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\deque(1601) : while compiling class template member function 'std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>> std::deque<Contained,std::allocator<_Ty>>::erase(std::_Deque_const_iterator<std::_Deque_val<std::_Deque_simple_types<_Ty>>>,std::_Deque_const_iterator<std::_Deque_val<std::_Deque_simple_types<_Ty>>>)'
1>          with
1>          [
1>              _Ty=Contained
1>          ]
1>          ConsoleApplication13.cpp(27) : see reference to function template instantiation 'std::_Deque_iterator<std::_Deque_val<std::_Deque_simple_types<Contained>>> std::deque<Contained,std::allocator<_Ty>>::erase(std::_Deque_const_iterator<std::_Deque_val<std::_Deque_simple_types<_Ty>>>,std::_Deque_const_iterator<std::_Deque_val<std::_Deque_simple_types<_Ty>>>)' being compiled
1>          with
1>          [
1>              _Ty=Contained
1>          ]
1>          ConsoleApplication13.cpp(31) : see reference to class template instantiation 'std::deque<Contained,std::allocator<_Ty>>' being compiled
1>          with
1>          [
1>              _Ty=Contained
1>          ]

Answer 1

简短回答：因为

Type requirements

-T must meet the requirements of MoveAssignable.

长答案：即使 std::deque 不需要提供连续内存，它仍然需要提供恒定复杂性 operator[].这就是它必须移动元素的原因。

Answer 2

这是因为 std::deque 通常实现为环形缓冲区，而环形缓冲区又通常实现为单件内存缓冲区。这意味着当您从双端队列中删除元素时，如果删除的元素既不在序列的末尾也不在序列的开头，则可能需要移动一些元素。这是插图：

    V buffer begins here                    V buffer ends here
1. [ ] [.] [.] [.] [.] [.] [.] [.] [ ] [ ] [ ]
        ^first element          ^last element

2. [ ] [.] [.] [.] [.] [.] [.] [.] [ ] [ ] [ ]
                ^ you want to remove this element.

                <= these elements should be moved
                    V   V   V   V
3. [ ] [.] [.] [ ] [:] [:] [:] [:] [ ] [ ] [ ]
                ^ element have been removed.

实际上赋值运算符仅在类型没有移动运算符时才使用。因此，如果您将以下行添加到 class，那么一切都可以正常编译：

Contained& operator=(Contained&&) = default;

更新： 看来我错了，因为大多数 STL 实现现在使用 Dynamic Arrays 的一些变体，而不是环形缓冲区。尽管如此，它们仍然是数组，如果从数组中间移除一个元素，则需要移动元素。

为什么 deque::erase() 调用赋值运算符？

Why does deque::erase() invoke assignment operator?

c++

c++11

deque