[英]How do I convert iterator to const_iterator in my custom list iterator class?
我正在嘗試在C ++中實現自己的雙向鏈接列表(my_list)代碼,尤其是列表的迭代器類。 我的問題是我想從迭代器到const_iterator進行隱式轉換,例如代碼my_list::iterator it = l.begin();
其中l
是my_list
編譯的實例。 但是,我沒有找到編譯器沒有抱怨的方法。
這是實現列表節點和迭代器類的代碼:
template<class T> class node {
node(const T& t = T()):data(t),next(0),prev(0) {}
T data;
node* next;
node* prev;
friend class my_list<T>;
friend class my_list_iterator<T>;
};
template<class T> class my_list_iterator {
public:
// increment and decrement operators
my_list_iterator operator++();
my_list_iterator operator++(int);
my_list_iterator operator--();
my_list_iterator operator--(int);
// bool comparison iterators
bool operator==(const my_list_iterator& other) const {return pos_==other.pos_;}
bool operator!=(const my_list_iterator& other) const {return pos_!=other.pos_;}
// member access
T& operator*() const {return pos_->data;}
T* operator->() const {return &(pos_->data);}
// implicit conversion to const iterator
operator my_list_iterator<const T>() {return my_list_iterator<const T>(pos_);}
private:
node<T>* pos_;
explicit my_list_iterator(node<T>* p=0):pos_(p) {}
friend class my_list<T>;
};
我已經省略了my_list實現,但是如果您認為它是相關的,則可以包括它。 當我測試此代碼時,它不會在GCC上編譯並出現以下錯誤:
In file included from test.cpp:2:
my_list.h: In instantiation of ‘my_list_iterator<T>::operator my_list_iterator<const T>() [with T = int]’:
test.cpp:12:49: required from here
my_list.h:37:48: error: no matching function for call to ‘my_list_iterator<const int>::my_list_iterator(node<int>*&)’
operator my_list_iterator<const T>() {return my_list_iterator<const T>(pos_);}
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
my_list.h:40:12: note: candidate: ‘my_list_iterator<T>::my_list_iterator(node<T>*) [with T = const int]’
explicit my_list_iterator(node<T>* p=0):pos_(p) {}
^~~~~~~~~~~~~~~~
my_list.h:40:12: note: no known conversion for argument 1 from ‘node<int>*’ to ‘node<const int>*’
my_list.h:20:25: note: candidate: ‘constexpr my_list_iterator<const int>::my_list_iterator(const my_list_iterator<const int>&)’
template<class T> class my_list_iterator {
^~~~~~~~~~~~~~~~
my_list.h:20:25: note: no known conversion for argument 1 from ‘node<int>*’ to ‘const my_list_iterator<const int>&’
my_list.h:20:25: note: candidate: ‘constexpr my_list_iterator<const int>::my_list_iterator(my_list_iterator<const int>&&)’
my_list.h:20:25: note: no known conversion for argument 1 from ‘node<int>*’ to ‘my_list_iterator<const int>&&’
有人可以幫我解決我做錯的事情嗎?
my_list.h:40:12: note: candidate: ‘my_list_iterator<T>::my_list_iterator(node<T>*) [with T = const int]’
explicit my_list_iterator(node<T>* p=0):pos_(p) {}
^~~~~~~~~~~~~~~~
my_list.h:40:12: note: no known conversion for argument 1 from ‘node<int>*’ to ‘node<const int>*’
node<int>
和node<const int>
是不相關的類型。 您不能將指向node<int>
的指針傳遞給需要指向node<const int>
的指針的node<const int>
。
除了可以在包含的類型上模板迭代器類之外,還可以將const
上移一層並在節點類型上模板化迭代器:
template<class Node> class my_list_iterator {
public:
//...
// member access
auto& operator*() const {return pos_->data;}
auto* operator->() const {return &(pos_->data);}
// implicit conversion to const iterator
operator my_list_iterator<const Node>() {return my_list_iterator<const Node>{pos_};}
private:
Node* pos_;
explicit my_list_iterator(Node* p=0):pos_(p) {}
friend class my_list<type>;
};
template <class T> class my_list {
public:
using iterator = my_list_iterator<node<T>>;
using const_iterator = my_list_iterator<const node<T>>;
//...
};
現在,您將一個指向node<int>
的指針傳遞給一個期望指向const node<int>
的指針的const node<int>
,這很好。
這是一種方法:
template <typename T>
class my_list {
public:
using iterator = my_list_iterator<T>;
using const_iterator = my_list_iterator<const T>;
const_iterator cbegin() const { return {/*...*/}; }
const_iterator cend() const { return {/*...*/}; }
const_iterator begin() const { return {/*...*/}; }
const_iterator end() const { return {/*...*/}; }
iterator begin() { return {/*...*/}; }
iterator end() { return {/*...*/}; }
};
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