4

我很想 通过以下方式解决这个问题:

#include <iostream>
#include <set>
#include <iterator>
#include <array>
#include <tuple>
#include <type_traits>

int main()
{
    const std::set<int> s{ 0, 1, 2, 3, 4, 5, 6, 7, 8 };
    auto iter = s.find(5);
    using IterType = decltype(iter);

    // using `std::array` works fine!
    const auto& [pv1, nxt1] = std::array<IterType, 2>{std::prev(iter), std::next(iter)};
    std::cout <<"using std::array<IterType, 2> :"<< *pv1 << " " << *nxt1 << '\n'; // prints: 4 6

    // using ` std::make_tuple` works fine!
    const auto& [pv2, nxt2] = std::make_tuple(std::prev(iter), std::next(iter));
    std::cout << "using std::make_tuple :" << *pv2 << " " << *pv2 << '\n';        // prints: 4 6

    // using `std::tie` deduction happens in MSVC, but not in GCC and Clang
    const auto& [pv3, nxt3] = std::tie(std::prev(iter), std::next(iter));
    // following is an assertion failure in MSVC with /O2  /std:c++17
    std::cout << "using std::tie :" << *pv3 << " " << *nxt3<< '\n';
}

I std::tied 返回的迭代器std::prevand std::next,并允许结构化绑定进行auto推导。

const auto& [pv3, nxt3] = std::tie(std::prev(iter), std::next(iter));

看起来它允许的唯一编译器是MSVC v19.14/O2 /std:c++17GCC 9.1clang 8.0不同意这一点。见在线编译器:https ://godbolt.org/z/DTb_OZ

海湾合作委员会 说:

<source>:23:28: error: no matching function for call to 'tie'
        const auto& [pv3, nxt3] = std::tie(std::prev(iter), std::next(iter));
                                  ^~~~~~~~
/opt/compiler-explorer/gcc-8.3.0/lib/gcc/x86_64-linux-gnu/8.3.0/../../../../include/c++/8.3.0/tuple:1605:5: note: candidate function [with _Elements = <std::_Rb_tree_const_iterator<int>, std::_Rb_tree_const_iterator<int>>] not viable: expects an l-value for 1st argument
    tie(_Elements&... __args) noexcept
^

铿锵声 说:

<source>: In function 'int main()':
<source>:23:46: error: cannot bind non-const lvalue reference of type 'std::_Rb_tree_const_iterator<int>&' to an rvalue of type 'std::_Rb_tree_const_iterator<int>'
   23 |  const auto& [pv3, nxt3] = std::tie(std::prev(iter), std::next(iter));
      |                                     ~~~~~~~~~^~~~~~
In file included from <source>:5:
/opt/compiler-explorer/gcc-9.1.0/include/c++/9.1.0/tuple:1611:19: note:   initializing argument 1 of 'constexpr std::tuple<_Elements& ...> std::tie(_Elements& ...) [with _Elements = {std::_Rb_tree_const_iterator<int>, std::_Rb_tree_const_iterator<int>}]'
 1611 |     tie(_Elements&... __args) noexcept
      |         ~~~~~~~~~~^~~~~~~~~~

查看cppreference.com 中给出的示例 MSVC 是否正确?或谁在这里,为什么?

有趣的是,在运行

std::cout << "using std::tie :" << *pv3 << " " << *nxt3<< '\n';

给我

在此处输入图像描述

(在MSVS 2019中,/std:c++17

4

1 回答 1

5

每个[tuple.creation]/tie

template<class... TTypes>
  constexpr tuple<TTypes&...> tie(TTypes&... t) noexcept;

在这里,参数是非常量左值引用。 std::prev(iter)并且std::next(iter)不能绑定到左值引用,所以代码应该被拒绝。MSVC 接受这一点的原因在评论中进行了解释:

启用/Za编译器标志,然后代码将被拒绝。MVSC [原文如此] 具有允许将临时绑定到左值引用的扩展。 tie获取左值引用,但是prevnext返回临时。– rafix07 2019-07-25 07:52:58Z

于 2019-07-25T08:53:00.357 回答