c++ - 如何使用 C++ 中的平凡仿函数避免 "invalid initialization of reference"错误

Question

我有一个“功能迭代器”类型的类——也就是说，迭代器返回的值首先由一个函数对象处理，它是模板参数。此模板参数的默认值为“平凡仿函数”，它只返回它收到的参数。

当我添加 const 成员函数时出现问题，该成员函数应返回对元素的 const 引用。为简单起见，假设我们有一个“前向迭代器”并希望有一个函数“peek_next”，它允许查看下一个元素，但不能更改它。这是我的看法(这是最小的可编译示例):

#include <functional>
#include <iostream>
#include <vector>
/**
 * This is an empty functor which just returns back its argument.
 */
template <class T>
class basic_functor: public std::unary_function<T, T>
{
public:
  typedef basic_functor self;
  basic_functor()
  {}

  /**
   * Returns back the argument passed to it. const version
   */
   inline const T& operator()(const T& arg) const
   { return arg; }

   /**
    * Returns back the argument passed to it.
    */
   inline T& operator()(T& arg)
   { return arg; }

};// class basic_functor

/**
 * @param S type of the elements.
 * @param BaseIterator type of the iterator it builds upon.
 * @param functor which is applied to the element to which iterator points.
 *      by default it is empty functor -- the element itself is returned.
 */
template <typename S,
          typename BaseIterator,
          typename Functor=basic_functor<S> >
class iterator_functional:
  public std::iterator<std::forward_iterator_tag,
                          typename Functor::result_type >
{
public:
  ///type defining itself
  typedef iterator_functional self;
  ///type of the functor
  typedef Functor functor_type;
  ///type of linear iterator
  typedef BaseIterator base_iterator_type;
  ///type for rebinding the iterator with another functor
  template <typename F>
  struct rebind
  {
    typedef iterator_functional<S, base_iterator_type, F> other;
  };

  /**
   * Constructor. 
   */
  iterator_functional(base_iterator_type it,
                       functor_type funct = functor_type()):
    m_it(it),
    m_functor(funct)
    {
    }

   /**
    * Copying constructor.
    * @param other source of the data.
    */
   iterator_functional(const self&other):
     m_it(other.m_it),
     m_functor(other.m_functor)
     { }

  /**
   * Provides access to element to which iterator points.
   * @return element to which iterator points to.
   */
  inline typename self::reference operator*()
    { return m_functor(*m_it); }

  /**
   * Provides access to element to which iterator points.
   * @return element to which iterator points to.
   */
  inline const typename self::reference operator*() const
    { return m_functor(*m_it); }
// Will work only when the code below is 'unlocked'
#if 0
  /**
   * Returns reference to the next element.
   */
  inline typename self::reference peek_next() 
  {
    auto temp_it = m_it;
    ++temp_it;
    return
       m_functor(*temp_it);
  }//iterator_type_const
#endif

  /**
   * Returns reference to the next element.
   */
  inline const typename self::reference peek_next() const
  {
    auto temp_it = m_it;
    ++temp_it;
    return
       m_functor(*temp_it);
  }//iterator_type_const

private:
  base_iterator_type m_it;
  functor_type m_functor;
};//iterator_functional;

typedef std::vector<int>::iterator v_int_iterator;
typedef std::vector<int>::const_iterator v_int_const_iterator;

typedef iterator_functional<int, v_int_iterator> iterator_type;
typedef iterator_functional<const int, v_int_const_iterator> iterator_type_const;

int main()
{
  std::vector<int> vec_1({1, 2, 3, 4});
  const std::vector<int> vec_2({5, 6, 7, 8});

  iterator_type it(vec_1.begin());
  std::cout << " *it =" << *it << ";  it.peek_next() = " << it.peek_next() << std::endl;

  iterator_type_const cit(vec_2.begin());
  std::cout << "*cit =" << *cit << "; cit.peek_next() = " << cit.peek_next() << std::endl;
  return 0;
}

问题是要编译此示例，我必须允许非 const 版本的 peek_element 函数。否则编译器失败并显示消息

error: invalid initialization of reference of type 'std::iterator<std::forward_iterator_tag, int, long int, int*, int&>::reference {aka int&}' from expression of type 'const int'

据我所知，编译器使用 basic_functor::operator() 的非常量版本。

那么我该如何避免呢？是的，我知道 unary_function 已从 C++17 中删除。

Answer 1

好的，感谢评论中的讨论(@danadam 和@DavisHerring)，我找到了答案。我误读了错误。编译器使用了正确版本的 basic_functor::operator() 好吧——问题是 iterator_functional::peek_next() 的返回类型。

我认为 const typename self::reference 与 const S& 相同，但事实并非如此。这不是对 const 的引用，而是 const 引用!因此，const 位被忽略，因此我试图将 basic_functor::operator() 返回的 const S& 转换为 S& -- iterator_functional::peek_next() 的返回类型。

因此，解决方案不是使用const typename self::reference，而是

  inline const typename self::value_type& operator*() const
    { return m_functor(*m_it); }

相反。

谁知道!

附言在与@DavisHerring 进一步讨论后，我将改进后的工作示例放在这里。亮点:

1. basic_functor 不需要第二个 operator()。只要是用const T模板参数创建即可。
2. 我已删除 S 模板参数并将其从 BaseIterator 中获取。请注意，我需要将 std::remove_reference 与 BaseIterator::reference 一起使用，因为 const_iterator::value_type 不是常量类型。<

所以在这里(ta-da):

#include <functional>
#include <iostream>
#include <vector>
/**
 * This is an empty functor which just returns back its argument.
 */
template <class T>
class basic_functor: public std::unary_function<T, T>
{
public:
  typedef basic_functor self;
  basic_functor()
  {}

   /**
    * Returns back the argument passed to it.
    */
   inline T& operator()(T& arg) const
   { return arg; }

};// class basic_functor

/**
 * @param S type of the elements.
 * @param BaseIterator type of the iterator it builds upon.
 * @param functor which is applied to the element to which iterator points.
 *      by default it is empty functor -- the element itself is returned.
 */
template <typename BaseIterator,
          typename Functor=basic_functor<
             typename std::remove_reference<
                   typename BaseIterator::reference>::type > >
class iterator_functional:
  public std::iterator<std::forward_iterator_tag,
                       typename Functor::result_type>
{
public:
  ///type defining itself
  typedef iterator_functional self;
  ///type of the functor
  typedef Functor functor_type;
  ///type of linear iterator
  typedef BaseIterator base_iterator_type;

  /**
   * Constructor. 
   */
  iterator_functional(base_iterator_type it,
                       functor_type funct = functor_type()):
    m_it(it),
    m_functor(funct)
    {
    }

   /**
    * Copying constructor.
    * @param other source of the data.
    */
   iterator_functional(const self&other):
     m_it(other.m_it),
     m_functor(other.m_functor)
     { }

  /**
   * Provides access to element to which iterator points.
   * @return element to which iterator points to.
   */
  inline typename self::reference operator*() const
    { return m_functor(*m_it); }

  /**
   * Returns reference to the next element.
   */
  inline const typename self::value_type& peek_next() const
  {
    auto temp_it = m_it;
    ++temp_it;
    return
       m_functor(*temp_it);
  }//iterator_type_const

private:
  base_iterator_type m_it;
  functor_type m_functor;
};//iterator_functional;

typedef std::vector<int>::iterator v_int_iterator;
typedef std::vector<int>::const_iterator v_int_const_iterator;

typedef iterator_functional<v_int_iterator> iterator_type;
typedef iterator_functional<v_int_const_iterator> iterator_type_const;

int main()
{
  std::vector<int> vec_1({1, 2, 3, 4});
  const std::vector<int> vec_2({5, 6, 7, 8});

  iterator_type it(vec_1.begin());
  *it = 9;
  std::cout << " *it =" << *it << ";  it.peek_next() = " << it.peek_next() << std::endl;

  iterator_type_const cit(vec_2.begin());
  std::cout << "*cit =" << *cit << "; cit.peek_next() = " << cit.peek_next() << std::endl;
  return 0;
}