我正在开发一个使用结构的库，该结构不应具有该库用户可访问的默认构造函数。

struct Example
{
    Example(int x);

  private:
    Example();
};

在库中，std::map 需要默认构造函数来创建新条目。该库非常小心地在使用默认构造函数的任何地方实际放置值。

库使用映射来存储这些结构，如下所示:

std::map<int, Example> data;

检查 HERE FOR A COMPLETE EXAMPLE在 ideOne 中。

我想阻止库的用户使用默认构造函数。我如何与 std::map、std::pair 和/或 std::tuple 交 friend 以允许 std::map 使用此默认构造函数？

friend class std::map<int, Example>;
friend class std::pair<int, Example>;

不工作，我不确定如何与以下构造函数交 friend ，它提示 beeing 无法访问默认构造函数:

// TEMPLATE CONSTRUCTOR pair::pair(tuple, tuple, sequence, sequence)
template<class _Ty1,
class _Ty2>
template<class _Tuple1,
    class _Tuple2,
    size_t... _Indexes1,
    size_t... _Indexes2> inline
    pair<_Ty1, _Ty2>::pair(_Tuple1& _Val1,
        _Tuple2& _Val2,
        integer_sequence<size_t, _Indexes1...>,
        integer_sequence<size_t, _Indexes2...>)
    : first(_STD get<_Indexes1>(_STD move(_Val1))...),
        second(_STD get<_Indexes2>(_STD move(_Val2))...)
    {   // construct from pair of tuples
    (void) _Val1;   // TRANSITION, VSO#181496
    (void) _Val2;
    }

非常感谢任何支持!

编辑:完整错误信息:

In file included from /usr/include/c++/6/bits/stl_map.h:63:0,
                 from /usr/include/c++/6/map:61,
                 from prog.cpp:2:
/usr/include/c++/6/tuple: In instantiation of ‘std::pair<_T1, _T2>::pair(std::tuple<_Args1 ...>&, std::tuple<_Args2 ...>&, std::_Index_tuple<_Indexes1 ...>, std::_Index_tuple<_Indexes2 ...>) [with _Args1 = {int&&}; long unsigned int ..._Indexes1 = {0ul}; _Args2 = {}; long unsigned int ..._Indexes2 = {}; _T1 = const int; _T2 = Example]’:
/usr/include/c++/6/tuple:1579:63:   required from ‘std::pair<_T1, _T2>::pair(std::piecewise_construct_t, std::tuple<_Args1 ...>, std::tuple<_Args2 ...>) [with _Args1 = {int&&}; _Args2 = {}; _T1 = const int; _T2 = Example]’
/usr/include/c++/6/ext/new_allocator.h:120:4:   required from ‘void __gnu_cxx::new_allocator<_Tp>::construct(_Up*, _Args&& ...) [with _Up = std::pair<const int, Example>; _Args = {const std::piecewise_construct_t&, std::tuple<int&&>, std::tuple<>}; _Tp = std::_Rb_tree_node<std::pair<const int, Example> >]’
/usr/include/c++/6/bits/alloc_traits.h:455:4:   required from ‘static void std::allocator_traits<std::allocator<_CharT> >::construct(std::allocator_traits<std::allocator<_CharT> >::allocator_type&, _Up*, _Args&& ...) [with _Up = std::pair<const int, Example>; _Args = {const std::piecewise_construct_t&, std::tuple<int&&>, std::tuple<>}; _Tp = std::_Rb_tree_node<std::pair<const int, Example> >; std::allocator_traits<std::allocator<_CharT> >::allocator_type = std::allocator<std::_Rb_tree_node<std::pair<const int, Example> > >]’
/usr/include/c++/6/bits/stl_tree.h:543:32:   required from ‘void std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_M_construct_node(std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_Link_type, _Args&& ...) [with _Args = {const std::piecewise_construct_t&, std::tuple<int&&>, std::tuple<>}; _Key = int; _Val = std::pair<const int, Example>; _KeyOfValue = std::_Select1st<std::pair<const int, Example> >; _Compare = std::less<int>; _Alloc = std::allocator<std::pair<const int, Example> >; std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_Link_type = std::_Rb_tree_node<std::pair<const int, Example> >*]’
/usr/include/c++/6/bits/stl_tree.h:560:4:   required from ‘std::_Rb_tree_node<_Val>* std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_M_create_node(_Args&& ...) [with _Args = {const std::piecewise_construct_t&, std::tuple<int&&>, std::tuple<>}; _Key = int; _Val = std::pair<const int, Example>; _KeyOfValue = std::_Select1st<std::pair<const int, Example> >; _Compare = std::less<int>; _Alloc = std::allocator<std::pair<const int, Example> >; std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_Link_type = std::_Rb_tree_node<std::pair<const int, Example> >*]’
/usr/include/c++/6/bits/stl_tree.h:2196:64:   required from ‘std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::iterator std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::_M_emplace_hint_unique(std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::const_iterator, _Args&& ...) [with _Args = {const std::piecewise_construct_t&, std::tuple<int&&>, std::tuple<>}; _Key = int; _Val = std::pair<const int, Example>; _KeyOfValue = std::_Select1st<std::pair<const int, Example> >; _Compare = std::less<int>; _Alloc = std::allocator<std::pair<const int, Example> >; std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::iterator = std::_Rb_tree_iterator<std::pair<const int, Example> >; std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::const_iterator = std::_Rb_tree_const_iterator<std::pair<const int, Example> >]’
/usr/include/c++/6/bits/stl_map.h:502:8:   required from ‘std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](std::map<_Key, _Tp, _Compare, _Alloc>::key_type&&) [with _Key = int; _Tp = Example; _Compare = std::less<int>; _Alloc = std::allocator<std::pair<const int, Example> >; std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type = Example; std::map<_Key, _Tp, _Compare, _Alloc>::key_type = int]’
prog.cpp:17:8:   required from here
/usr/include/c++/6/tuple:1590:70: error: ‘Example::Example()’ is private within this context
         second(std::forward<_Args2>(std::get<_Indexes2>(__tuple2))...)
                                                                      ^
prog.cpp:11:2: note: declared private here
  Example(){}
  ^~~~~~~

最佳答案

首先，我想纠正您帖子中的一个重点:

Inside the library the default constructor is required for std::map

您可以使用 std::map<K,T>即使 T 没有默认构造函数。参见 this post .在这种情况下，您不能使用 operator[]读取和写入 map 。您仍然可以使用其他方法来完成:

• 阅读 map :V value = map.at(key);
• 插入 map :map.insert(std::make_pair(key, value)); .

我强烈建议这样做。

话虽这么说，如果你真的想从你的错误信息中走“私有(private)构造函数和 friend ”的方式:

In instantiation of ‘std::pair<_T1, _T2>::pair(/*...*/) [with /*...*/; _T1 = const int; _T2 = Example]’

你可以试试friend std::pair<const int, Example>; .正如 Caleth 在他的回答中所说，这可能不可移植。

关于c++ - 更正友元定义以授予 std::map 对私有(private)默认构造函数的访问权限，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50485678/