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c++ - 使用 std::pair 作为 operator= 的参数时的编译问题

问题描述

我被困了几个小时，试图弄清楚为什么以下代码无法编译，如果有人能指出我所缺少的，我将不胜感激。

代码只是我在真实项目代码中遇到的编译问题的简化示例。

#include <map>
#include <string>
#include <utility>

class A
{
public:
   A() = default;
   A(const A& v) = default;
   A(A&& v) = default;
   A& operator=(const A& v) = default;
   
   A& operator=(const std::pair<A, A>& v)
   {
      return *this;
   }
};

void func(const std::pair<int, A>& obj);

int main(int argc, char *argv[])
{
    std::pair<int, A> obj;
    func(obj);
    return 0;
}

void func(const std::pair<int, A>& obj)
{
    A a, b;
    a = b;
}

该问题似乎与"A& operator=(const std::pair<A, A>& v)"相关。让我感到困惑的是，如果我在main()之前移动函数“void func(...)” ，一切都会正确编译。

使用 gcc (C++17) 编译。我得到的错误如下所示：

In file included from /usr/include/c++/7/bits/move.h:54:0,
                 from /usr/include/c++/7/bits/stl_pair.h:59,
                 from /usr/include/c++/7/bits/stl_algobase.h:64,
                 from /usr/include/c++/7/bits/stl_tree.h:63,
                 from /usr/include/c++/7/map:60,
                 from main.cpp:1:
/usr/include/c++/7/type_traits: In instantiation of ‘struct std::__and_<std::is_copy_assignable<A>, std::is_copy_assignable<A> >’:
/usr/include/c++/7/bits/stl_pair.h:378:7:   required from ‘struct std::pair<A, A>’
/usr/include/c++/7/type_traits:1259:45:   required by substitution of ‘template<class _Tp1, class _Up1, class> static std::true_type std::__is_assignable_helper<A&, const A&>::__test<_Tp1, _Up1, <template-parameter-1-3> >(int) [with _Tp1 = A&; _Up1 = const A&; <template-parameter-1-3> = <missing>]’
/usr/include/c++/7/type_traits:1268:40:   required from ‘class std::__is_assignable_helper<A&, const A&>’
/usr/include/c++/7/type_traits:1273:12:   required from ‘struct std::is_assignable<A&, const A&>’
/usr/include/c++/7/type_traits:1285:12:   required from ‘struct std::__is_copy_assignable_impl<A, true>’
/usr/include/c++/7/type_traits:1291:12:   required from ‘struct std::is_copy_assignable<A>’
/usr/include/c++/7/type_traits:143:12:   required from ‘struct std::__and_<std::is_copy_assignable<int>, std::is_copy_assignable<A> >’
/usr/include/c++/7/bits/stl_pair.h:378:7:   required from ‘struct std::pair<int, A>’
<span class="error_line" onclick="ide.gotoLine('main.cpp',23)">main.cpp:23:20</span>:   required from here
/usr/include/c++/7/type_traits:143:12: error: incomplete type ‘std::is_copy_assignable’ used in nested name specifier
     struct __and_<_B1, _B2>
            ^~~~~~~~~~~~~~~~
In file included from /usr/include/c++/7/bits/stl_algobase.h:64:0,
                 from /usr/include/c++/7/bits/stl_tree.h:63,
                 from /usr/include/c++/7/map:60,
                 from main.cpp:1:
/usr/include/c++/7/bits/stl_pair.h: In instantiation of ‘struct std::pair<A, A>’:
/usr/include/c++/7/type_traits:1259:45:   required by substitution of ‘template<class _Tp1, class _Up1, class> static std::true_type std::__is_assignable_helper<A&, const A&>::__test<_Tp1, _Up1, <template-parameter-1-3> >(int) [with _Tp1 = A&; _Up1 = const A&; <template-parameter-1-3> = <missing>]’
/usr/include/c++/7/type_traits:1268:40:   required from ‘class std::__is_assignable_helper<A&, const A&>’
/usr/include/c++/7/type_traits:1273:12:   required from ‘struct std::is_assignable<A&, const A&>’
/usr/include/c++/7/type_traits:1285:12:   required from ‘struct std::__is_copy_assignable_impl<A, true>’
/usr/include/c++/7/type_traits:1291:12:   required from ‘struct std::is_copy_assignable<A>’
/usr/include/c++/7/type_traits:143:12:   required from ‘struct std::__and_<std::is_copy_assignable<int>, std::is_copy_assignable<A> >’
/usr/include/c++/7/bits/stl_pair.h:378:7:   required from ‘struct std::pair<int, A>’
<span class="error_line" onclick="ide.gotoLine('main.cpp',23)">main.cpp:23:20</span>:   required from here
/usr/include/c++/7/bits/stl_pair.h:378:7: error: ‘value’ is not a member of ‘std::__and_, std::is_copy_assignable >’
       operator=(typename conditional<
       ^~~~~~~~

标签： c++templates

解决方案

您现在所经历的是未定义行为的主要示例，因为std::pair实现取决于type_traits未为不完整类型定义的行为。

在任何成员函数的范围内A但在任何成员函数之外都A被认为是不完整的类型。您正在尝试定义一个接受 a 的函数std::pair<A, A>，因此std::pair使用不完整的类型进行实例化。

如果A是模板类型，则情况并非如此，因为类模板的函数在实际调用之前不会被实例化。

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