为什么我的自定义迭代器不适用于 STL 副本？

Question

我为answer to another question 编写了一个OutputIterator。这里是：

#include <queue>

using namespace std;

template< typename T, typename U >
class queue_inserter {
    queue<T, U> &qu;  
public:
    queue_inserter(queue<T,U> &q) : qu(q) { }
    queue_inserter<T,U> operator ++ (int) { return *this; }
    queue_inserter<T,U> operator * () { return *this; }
    void operator = (const T &val) { qu.push(val); }
};

template< typename T, typename U >
queue_inserter<T,U> make_queue_inserter(queue<T,U> &q) {
    return queue_inserter<T,U>(q);
}    

这非常适合这个小复制功能：

template<typename II, typename OI>
void mycopy(II b, II e, OI oi) {
    while (b != e) { *oi++ = *b++; }
}

但对于来自 algorithms 的 STL copy，它根本不起作用。以下是我得到的美妙的 C++ 错误：

i.cpp:33: error: specialization of ‘template<class _Iterator> struct std::iterator_traits’ in different namespace
/usr/include/c++/4.0.0/bits/stl_iterator_base_types.h:127: error:   from definition of ‘template<class _Iterator> struct std::iterator_traits’
/usr/include/c++/4.0.0/bits/stl_algobase.h: In function ‘_OI std::__copy_aux(_II, _II, _OI) [with _II = int*, _OI = queue_inserter<int, std::deque<int, std::allocator<int> > >]’:
/usr/include/c++/4.0.0/bits/stl_algobase.h:335:   instantiated from ‘static _OI std::__copy_normal<true, false>::copy_n(_II, _II, _OI) [with _II = __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >, _OI = queue_inserter<int, std::deque<int, std::allocator<int> > >]’
/usr/include/c++/4.0.0/bits/stl_algobase.h:387:   instantiated from ‘_OutputIterator std::copy(_InputIterator, _InputIterator, _OutputIterator) [with _InputIterator = __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >, _OutputIterator = queue_inserter<int, std::deque<int, std::allocator<int> > >]’
i.cpp:53:   instantiated from here
/usr/include/c++/4.0.0/bits/stl_algobase.h:310: error: no type named ‘value_type’ in ‘struct std::iterator_traits<queue_inserter<int, std::deque<int, std::allocator<int> > > >’
/usr/include/c++/4.0.0/bits/stl_algobase.h:315: error: no type named ‘value_type’ in ‘struct std::iterator_traits<queue_inserter<int, std::deque<int, std::allocator<int> > > >’
/usr/include/c++/4.0.0/bits/stl_algobase.h:315: error: ‘__value’ is not a member of ‘<declaration error>’
/usr/include/c++/4.0.0/bits/stl_algobase.h:335:   instantiated from ‘static _OI std::__copy_normal<true, false>::copy_n(_II, _II, _OI) [with _II = __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >, _OI = queue_inserter<int, std::deque<int, std::allocator<int> > >]’
/usr/include/c++/4.0.0/bits/stl_algobase.h:387:   instantiated from ‘_OutputIterator std::copy(_InputIterator, _InputIterator, _OutputIterator) [with _InputIterator = __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >, _OutputIterator = queue_inserter<int, std::deque<int, std::allocator<int> > >]’
i.cpp:53:   instantiated from here
/usr/include/c++/4.0.0/bits/stl_algobase.h:317: error: ‘__simple’ is not a valid template argument for type ‘bool’ because it is a non-constant expression
/usr/include/c++/4.0.0/bits/stl_algobase.h:317: error: ‘copy’ is not a member of ‘<declaration error>’

这是驱动程序：

int main() {
    vector<int> v;
    v.push_back( 1 );
    v.push_back( 2 );
    queue<int> q;
    copy( v.begin(), v.end(), make_queue_inserter(q) );
    while (q.size() > 0) {
        cout << q.front() << endl;
        q.pop();
    }
}

为什么它专门针对iterator_traits。我的迭代器有什么问题？我不能只编写自己的简单迭代器吗？

Answer 1

您的queue_inserter 需要从std::iterator 派生，以便正确定义所有类型定义，例如value_type，因为它们在STL 算法中使用此定义有效：

template< typename T, typename U >
class queue_inserter : public std::iterator<std::output_iterator_tag, T>{
    queue<T, U> &qu;  
public:
    queue_inserter(queue<T,U> &q) : qu(q) { }
    queue_inserter<T,U> operator ++ (int) { return *this; }
    queue_inserter<T,U> operator ++ () { return *this; }
    queue_inserter<T,U> operator * () { return *this; }
    void operator = (const T &val) { qu.push(val); }
};

Answer 2

从 std::iterator 派生它。如果您有兴趣，Dobb 博士有一个关于自定义容器和迭代器的 article。

Answer 3

您的迭代器不满足“可分配”类型的要求，这是输出迭代器的要求，因为它包含引用，并且可分配类型需要确保在 t = u 之后 t 等于 @987654323 @。

您可以通过从 std::iterator 的特化派生或显式提供一个为您的迭代器提供合适的 iterator_traits 特化。

namespace std
{
    template<> struct iterator_traits<MyIterator>
    {
        typedef std::output_iterator_tag iterator_category;
        typedef void value_type;
        typedef void difference_type;
    };
}

Answer 4

#include <queue>
#include <algorithm>
#include <iterator>
#include <iostream>

using namespace std;

template< typename T, typename U >
class queue_inserter
{
    queue<T, U> &qu;

public:
    // for iterator_traits to refer
    typedef output_iterator_tag iterator_category;
    typedef T value_type;
    typedef ptrdiff_t difference_type;
    typedef T* pointer;
    typedef T& reference;

    queue_inserter(queue<T,U> &q) : qu(q) { }
    queue_inserter<T,U>& operator ++ () { return *this; }
    queue_inserter<T,U> operator * () { return *this; }
    void operator = (const T &val) { qu.push(val); }
};

template< typename T, typename U >
queue_inserter<T,U> make_queue_inserter(queue<T,U> &q)
{
    return queue_inserter<T,U>(q);
}

int main()
{
    // uses initalizer list (C++0x), pass -std=c++0x to g++
    vector<int> v({1, 2, 3});
    queue<int, deque<int>> q;
    copy(v.cbegin(), v.cend(), make_queue_inserter(q));
    while (!q.empty())
    {
        cout << q.front() << endl;
        q.pop();
    }
}

这应该用iterator_traits来做； <iterator> 中的辅助结构，它定义了迭代器通常应该定义的所有类型。 <algorithm> 中的函数，在需要时引用这些类型，例如 iterator_traits<it>::iterator_category 或说 iterator_traits<it>::value_type 等。只需在自定义迭代器中定义它们就可以了。这是编写迭代器的现代方式，而不是从std::iterator 继承的经典方式。查看<iterator> 会发现，即使std::iterator 也定义了这些类型，即iterator_category、difference_type 等。这就是从std::iterator 继承时，派生的迭代器类由于遗传而获得这些类型的原因。