std :: map的淺/深拷貝

Question

我如何最好地實現這些？我認爲是這樣的：

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)); 
    } 

成員data定義如下：

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

這不起作用，很遺憾。以下是編譯器錯誤，我真的不明白他們：

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，這看起來過於複雜我。 是真的，我可以做得更好嗎？

順便說一句，我在我從形狀派生的類的clone（）方法 。也許我可以將它們用於clone_deep方法 ？他們好嗎？他們看起來像這樣的東西 ：在你的方法

所有例子

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

通常一個克隆函數會返回一個指向新實例的指針。你正在返回的是一個按值創建的對象，它是從一個動態分配的實例中複製而來，然後被泄露。

如果您想通過要返回的值，那麼你不應該使用new。 例如

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

如果data成員只是一個std::map實例，那麼它將被複製爲您的淺克隆在任何情況下的一部分，所以也沒有必要做std::copy在深克隆的情況下，它沒有試圖做任何不同的事。

如果您想要做一張地圖的std::copy，您需要使用std::insert_iterator。

我想，這可能是更容易做到事後每個形狀的clone，雖然。

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

首先泄漏內存，因爲你new一個shape_container但隨後它被通過返回的值複製出來。與您的shape示例一樣，您應該返回指針。

編譯器錯誤期待，因爲它的抱怨它不能爲你生成一個賦值操作符以某種方式來複制有關。再次嘗試使用指針，該問題應該消失。

Answer 3

如果你做了深拷貝的地圖，那麼你必須創建一個新的具有深拷貝的元素的創建地圖。

想想引用計數方法，這將是更好的辦法。

Answer 4

一種選擇是包裝你的形狀類型在執行對象的深 副本類型：

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; 
}; 

然後構造這種類型的地圖：

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