std::map 的浅/深副本

Question

我如何最好地实施这些？我想到了这样的事情：

    using namespace std;

    shape_container
    shape_container::clone_deep () const
    {
        shape_container* ptr = new shape_container();
        copy( data.begin(), data.end(), (*ptr).begin() );
        return *ptr;
    }

    shape_container
    shape_container::clone_shallow () const
    {
        return *( new shape_container(*this) );
    }

成员 data 定义如下：

    std::map<std::string, shape*> data;

不幸的是，这不起作用。这是编译器错误，我不太理解它们：

    g++ -Wall -O2 -pedantic -I../../UnitTest++/src/ -I./libfglwin/include/ -I. -c shape_container.cpp -o shape_container.o
    /usr/include/c++/4.2.1/bits/stl_pair.h: In member function ‘std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>& std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>::operator=(const std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>&)’:
    /usr/include/c++/4.2.1/bits/stl_pair.h:69:   instantiated from ‘static _OI std::__copy<<anonymous>, <template-parameter-1-2> >::copy(_II, _II, _OI) [with _II = std::_Rb_tree_const_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, _OI = std::_Rb_tree_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, bool <anonymous> = false, <template-parameter-1-2> = std::bidirectional_iterator_tag]’
    /usr/include/c++/4.2.1/bits/stl_algobase.h:315:   instantiated from ‘_OI std::__copy_aux(_II, _II, _OI) [with _II = std::_Rb_tree_const_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, _OI = std::_Rb_tree_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >]’
    /usr/include/c++/4.2.1/bits/stl_algobase.h:340:   instantiated from ‘static _OI std::__copy_normal<<anonymous>, <anonymous> >::__copy_n(_II, _II, _OI) [with _II = std::_Rb_tree_const_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, _OI = std::_Rb_tree_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, bool <anonymous> = false, bool <anonymous> = false]’
    /usr/include/c++/4.2.1/bits/stl_algobase.h:401:   instantiated from ‘_OutputIterator std::copy(_InputIterator, _InputIterator, _OutputIterator) [with _InputIterator = std::_Rb_tree_const_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, _OutputIterator = std::_Rb_tree_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >]’
    shape_container.cpp:70:   instantiated from here
    /usr/include/c++/4.2.1/bits/stl_pair.h:69: error: non-static const member ‘const std::basic_string<char, std::char_traits<char>, std::allocator<char> > std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>::first’, can't use default assignment operator
    /usr/include/c++/4.2.1/bits/stl_algobase.h: In static member function ‘static _OI std::__copy<<anonymous>, <template-parameter-1-2> >::copy(_II, _II, _OI) [with _II = std::_Rb_tree_const_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, _OI = std::_Rb_tree_iterator<std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*> >, bool <anonymous> = false, <template-parameter-1-2> = std::bidirectional_iterator_tag]’:
    /usr/include/c++/4.2.1/bits/stl_algobase.h:268: note: synthesized method ‘std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>& std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>::operator=(const std::pair<const std::basic_string<char, std::char_traits<char>, std::allocator<char> >, shape*>&)’ first required here 

不知怎的，这对我来说看起来不必要地复杂。是那个 是的，我可以做得更好吗？

顺便说一句，我有 clone() 方法 在我从形状衍生的课程中。也许我可以使用它们 对于clone_deep方法？他们还好吗？他们看起来有些东西 像这样：

    class shape
    {
        public:
            /* Many methods. */
            virtual shape* clone () const = 0;

        protected:
            colorRGB    color_;
            std::string name_;
    };

    class triangle2d : public shape
    {
        public:
            /* Many methods. */
            triangle2d* clone() const;
        private:
            point3d a_, b_, c_;
    };

    triangle2d*
    triangle2d::clone() const
    {
        return new triangle2d(*this);
    } 

Answer 1

通常克隆函数会返回一个指向新实例的指针。您返回的是一个按值对象，该对象是从动态分配的实例中复制构造的，然后会泄漏。

如果你想按值返回，那么你不应该使用new。
例如，

shape_container shape_container::clone_shallow () const
{
    return *this;
}

如果 data 成员只是一个 std::map 实例，那么在任何情况下它将被复制为浅克隆的一部分，因此无需执行以下操作std::copy 在深度克隆的情况下，它不会尝试做任何不同的事情。

如果您想对映射进行std::copy，则需要使用std::insert_iterator。

不过，我认为事后对每个形状进行克隆可能会更容易。

例如

shape_container shape_container::clone_deep() const
{
    shape_container ret(*this);

    for (std::map<std::string, shape*>::iterator i = ret.data.begin(); i != ret.data.end(); ++i)
    {
        i->second = i->second->clone();
    }

    return ret;
}

Answer 2

首先，您的示例会泄漏内存，因为您在方法中new一个shape_container，但随后它通过返回值被复制出来。您应该像 shape 示例一样返回指针。

编译器错误看起来在某种程度上与复制有关，因为它抱怨它无法为您生成赋值运算符。再次尝试使用指针，这个问题就会消失。

Answer 3

如果您对地图进行深度复制，那么您必须使用深度复制的所有元素创建一个新的地图。

考虑一下引用计数方法，这将是更好的方法。

Answer 4

一种选择是将形状类型包装在执行深度的类型中
对象的副本：

class shape_deep_copy_wrapper {
  // ...
public:
  shape_deep_copy_wrapper (shape * shape)
  : m_my_shape (shape)
  {
  }

  shape_deep_copy_wrapper (shape_deep_copy_wrapper const & rhs)
  : m_my_shape (rhs.m_my_shape.deep_copy ())
  {
  }

  // ...

private:
  shape * m_my_shape;
};

然后构造一个具有以下类型的映射：

typedef std :: map < shape_deep_copy_wrapper , ... > DeepCopy ;
typedef std :: map < shape* , ... >                  ShallowCopy ;