更正友元定义,使std :: map访问私有默认构造函数

Question

我的工作,它使用应一个结构库不具有由LIB的用户访问默认的构造函数.

struct Example
{
    Example(int x);

  private:
    Example();
};

在库内部,std :: map需要默认构造函数来创建新条目.库在实际上将值放在使用默认构造函数的任何地方都非常小心.

该库使用映射来存储这些结构,如下所示:

std::map<int, Example> data;

在这里查看ideOne中的完整示例.

我想阻止lib的用户能够使用默认构造函数.我如何与std :: map,std :: pair和/或std :: tuple建立联系以允许std :: map使用这个默认构造函数？

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

不工作,我不知道如何与下面的抱怨beeing无法访问默认构造函数的构造函数交朋友:

// 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:完整错误消息:

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(){}
  ^~~~~~~

Answer 1

首先,我想纠正你帖子中的重点:

在库内部,std :: map需要默认构造函数

std::map<K,T>即使T没有默认构造函数,也可以使用.看这篇文章.在这种情况下,您无法使用operator[]读写地图.您仍然可以使用其他方法:

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

我强烈建议这样做.

话虽这么说,如果你真的想要按照"私有构造函数和朋友"的方式,从你的错误信息:

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

你可以尝试friend std::pair<const int, Example>;.正如Caleth在他的回答中所说,这可能不是便携式的.