C++在函数中从back_insert_iterator转换为迭代器

C++ cast from back_insert_iterator to iterator within functors

本文关键字：insert iterator 迭代器 转换 back 函数 C++      更新时间：2023-10-16

我正在尝试实现一种复制运算符。目的是有两个类：一个浏览容器，另一个用它做一些事情。browse类还(出于某种原因)在outout容器上维护一个迭代器，而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？

#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++编译失败，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> >’

这是问题的主要来源：std::back_insert_iterator< V<T> >和std::vector<T>::iterator在它们的继承树中没有直接关联：

  • std::vector<T>::iterator是__normal_iterator<T>(没有其他超类)(std::vector<T>::iterator的stl_vector.h和__normal_iterator的stl_iterator.h)
  • std::back_insert_iterator< V<T> >是一个迭代器(没有其他的超类)(std::back_insert_iterator看stl_iterator.h，std::iterator看stl_iterator_base_types.h)

它们不能向任何方向转换。

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

我认为std::advance( iterator, 1 )是指++iterator，它是迭代器进入下一个元素的标准方式。

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

第38行，i_in的类型应该是typename TIN::iterator，而不是TIN。Browseoperator()还必须返回out迭代器。

最后的代码如下：

#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++，它使C++错误更加清晰。

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