我一直在努力解决与开发模板类迭代器相关的问题。更具体地说,以正确的方式实现解除引用运算符(operator *()),以便模板类涵盖迭代器和const_iterator的情况。我确信我错过了一些明显的东西,但我看不到它。你能帮帮我吗?
假设我有以下模板类(Iterator.hpp
),我想用它来迭代将STL容器(std :: map,std :: vector)包装为私有成员的类的对象。 / p>
#include <iostream>
#include <iterator>
template<typename iterator_type>
class Iterator
{
public:
/** \brief Type to be returned when de-referencing the iterator*/
typedef typename std::iterator_traits<iterator_type>::value_type value_type;
/** \brief Constructor*/
inline Iterator(iterator_type i) : iterator(i) {}
/** \brief Dereference operator */
inline value_type& operator*() {return *iterator;}
inline const value_type& operator*() const {return *iterator;}
/** \brief Increment operator */
inline Iterator & operator++() {++iterator; return *this;}
/** \brief Inequality operator */
inline bool operator!=(const Iterator & right) const
{return iterator != right.iterator;}
/** \brief Inequality operator */
inline bool operator!=(const iterator_type & right) const
{return iterator != right;}
/** \brief Distance between iterators */
inline int operator-(const Iterator & right) const
{return std::distance(right.iterator, iterator);}
/** \brief Distance between iterators */
inline int operator-(const iterator_type & right) const
{return std::distance(right, iterator);}
private:
/** \brief Internal member, of iterator type*/
iterator_type iterator;
};
现在假设我有一个主(example.cpp
)创建一个int向量并尝试使用const_iterator
打印出来,如下所示:
#include "Iterator.hpp"
#include <vector>
typedef std::vector<int> IntVector;
int main(int argc, char* argv[]) {
unsigned int nElements(10);
IntVector intVector(nElements);
for (unsigned int i = 0; i < nElements; ++i) {
intVector[i] = i;
}
std::cerr << " Printout of the vector \n";
Iterator<IntVector::const_iterator> it(intVector.begin());
for(; it != intVector.end(); ++it) {
std::cerr << *it << "\n";
}
}
如果我尝试编译此代码:g++ example.cpp -std=c++11 -stdlib=libc++
。我会收到以下错误:
./Iterator.hpp:16:48: error: binding of reference to type 'value_type' (aka 'int') to a value of type 'const int' drops
qualifiers
inline value_type& operator*() {return *iterator;}
^~~~~~~~~
example.cpp:19:22: note: in instantiation of member function 'Iterator<std::__1::__wrap_iter<const int *> >::operator*'
requested here
std::cerr << *it << "\n";
^
1 error generated.
我认为通过简单地使用operator*()
模板中的const和非const版本重载Iterator.hpp
就足够了,但显然情况并非如此。你知道我在这里缺少什么吗?对此问题的任何帮助将不胜感激。
非常感谢!
答案 0 :(得分:0)
const_iterator
和iterator
是不同的类型,应该作为单独的类型实现。 const_iterator应该在它的dereference运算符中返回一个const引用,normal iterator返回可修改的引用。
您不应该将const_iterator
与const iterator
混淆。