Question

我有点想要实现一种复制操作符。目标是有两个类：一个用于浏览容器，另一个用于处理容器。 Browse类还维护（出于某种原因）输出容器上的迭代器，另一个可以用它计算增量。

不幸的是，编译器似乎无法将back_insert_iterator转换为输出迭代器。为什么呢？

#include <iostream>
#include <iterator>
#include <vector>

typedef std::vector<int> Vec;

// An operator that copy an item onto another
template< class TIN, class TOUT >
class DoCopy
{
    TOUT operator()( const typename TIN::iterator i_in, const typename TOUT::iterator i_out )
    {
        const typename TOUT::iterator i_incr = i_out;
        (*i_incr) = (*i_in);
        std::advance( i_incr, 1 );
        return i_incr;
    }
};

// The class that iterate over a container, calling an operator for each item
template< class TIN, class TOUT >
class Browse
{
    public:
        // We keep a reference to the operator that really do the job
        DoCopy<TIN,TOUT> & _do;
        Browse( DoCopy<TIN,TOUT> & op ) : _do(op) {}

        // Iterate over an input container
        TOUT operator()(
                const typename TIN::iterator in_start,
                const typename TIN::iterator in_end,
                const typename TOUT::iterator out_start
            )
        {
            TOUT i_out = out_start;

            for( TIN i_in = in_start; i_in != in_end; ++i_in ) {
                // it is not shown why here, but we DO want the operator to increment i_out
                i_out = _do(i_in, i_out);
            }
        }
};

int main()
{
    // in & out could be the same type or a different one
    Vec in;
    Vec out;
    DoCopy<Vec,Vec> do_copy;
    Browse<Vec,Vec> copy(do_copy);

    std::back_insert_iterator< Vec > insert_back(out);

    // Here, g++ cannot find the corresponding function :
    copy( in.begin(), in.end(), insert_back );

}

g ++无法编译时出现以下错误：

$ g++ test.cpp && ./a.out
    test.cpp: In function ‘int main()’:
    test.cpp:54:49: erreur: no match for call to ‘(Browse<std::vector<int>, std::vector<int> >) (std::vector<int>::iterator, std::vector<int>::iterator, std::back_insert_iterator<std::vector<int> >&)’
    test.cpp:22:11: note: candidate is:
    test.cpp:30:18: note: TOUT Browse<TIN, TOUT>::operator()(typename TIN::iterator, typename TIN::iterator, typename TOUT::iterator) [with TIN = std::vector<int>, TOUT = std::vector<int>, typename TIN::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >, typename TOUT::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >]
    test.cpp:30:18: note:   no known conversion for argument 3 from ‘std::back_insert_iterator<std::vector<int> >’ to ‘__gnu_cxx::__normal_iterator<int*, std::vector<int> >’

Answer 1

以下是问题的主要来源：std::back_insert_iterator< V<T> >和std::vector<T>::iterator在其继承树中没有直接关联：

std::vector<T>::iterator是__normal_iterator<T>（没有其他超类）（stl_vector.h std::vector<T>::iterator和stl_iterator.h __normal_iterator
std::back_insert_iterator< V<T> >是一个迭代器（没有其他超类）（查看stl_iterator.h std::back_insert_iterator和stl_iterator_base_types.h std::iterator。）

它们无法向任何方向转换。

因此，第二个模板参数应该直接是std::back_insert_iterator或iterator<>，其中第一个参数指示是输出运算符。

通过std::advance( iterator, 1 )，我假设你的意思是++iterator，这是转到迭代器的下一个元素的标准方法。

此外，out迭代器不应该是const，否则它们不会实现效果operator=。

第38行，i_in应为typename TIN::iterator类型，而不是TIN。 Browse operator()还必须返回out迭代器。

最终代码如下：

#include <iostream>
#include <iterator>
#include <vector>

typedef std::vector<int> Vec;

// An operator that copy an item onto another
template< class TIN, class TOUT >
class DoCopy
{
    public:

    TOUT operator()( const typename TIN::iterator i_in, const TOUT i_out )
    {
        TOUT i_incr = i_out;
        (*i_incr) = (*i_in);
        //std::advance( i_incr, 1 );
        ++i_incr;
        return i_incr;
    }
};

// The class that iterate over a container, calling an operator for each item
template< class TIN, class TOUT >
class Browse
{
    public:
        // We keep a reference to the operator that really do the job
        DoCopy<TIN,TOUT> & _do;
        Browse( DoCopy<TIN,TOUT> & op ) : _do(op) {}

        // Iterate over an input container
        TOUT operator()(
                const typename TIN::iterator in_start,
                const typename TIN::iterator in_end,
                const TOUT out_start
            )
        {
            TOUT i_out = out_start;

            for( typename TIN::iterator i_in = in_start; i_in != in_end; ++i_in ) {
                // it is not shown why here, but we DO want the operator to increment i_out
                i_out = _do(i_in, i_out);
            }
            return i_out;
        }
};

int main()
{
    // in & out could be the same type or a different one
    Vec in;

    in.push_back(1);
    in.push_back(3);
    in.push_back(3);
    in.push_back(7);

    Vec out;
    DoCopy<Vec, std::back_insert_iterator<Vec> > do_copy;
    Browse<Vec, std::back_insert_iterator<Vec> > copy(do_copy);

    std::back_insert_iterator< Vec > insert_back(out);

    // Here, g++ cannot find the corresponding function :
    copy( in.begin(), in.end(), insert_back );

    for( unsigned i = 0, s = out.size(); i < s; ++i )
    {
        std::cout << out[i] << " ";
    }
    std::cout << std::endl;
}

感谢 clang ++ ，这使得C ++错误更加清晰。

C ++从back_insert_iterator转换为仿函数中的迭代器

1 个答案: