C++ std:：成员指针行为的元组

我想了解这种行为的原因：

此代码非常有效：

#include <iostream>
#include <tuple>

struct SomeClass {
 int v1;
 int v2;
 int v3;
};


auto pv1 = &SomeClass::v1;
auto pv2 = &SomeClass::v2;
auto pv3 = &SomeClass::v3;

auto t = std::tie(pv1, pv2, pv3);

int main() {

  SomeClass c;
  c.v1 = 111;
  c.v2 = 222;
  c.v3 = 333;

  std::cout << c.*std::get<0>(t) << std::endl << c.*std::get<1>(t) << std::endl << c.*std::get<2>(t) << std::endl;
}

#包括
#包括
结构类{
int v1；
int v2；
int v3；
};
自动pv1=&SomeClass:：v1；
自动pv2=&SomeClass:：v2；
自动pv3=&SomeClass:：v3；
自动t=std:：tie（pv1、pv2、pv3）；
int main（）{
c类；
c、 v1=111；
c、 v2=222；
c、 v3=333；
std：：cout如果我们查看CPP的参考文档，它说（我的重点）：

我们可以看到，将右值引用传递给
make\u tuple
确实有效：

模板
元组生成（类型和参数）；
^^

#include <iostream>
#include <tuple>

struct SomeClass {
 int v1;
 int v2;
 int v3;
};

auto t = std::tie(&SomeClass::v1, &SomeClass::v2, &SomeClass::v3);

int main() {

  SomeClass c;
  c.v1 = 111;
  c.v2 = 222;
  c.v3 = 333;

  std::cout << c.*std::get<0>(t) << std::endl << c.*std::get<1>(t) << std::endl << c.*std::get<2>(t) << std::endl;
}

test.cpp:17:65: error: invalid initialization of non-const reference of type ‘int SomeClass::*&’ from an rvalue of type ‘int SomeClass::*’
 auto t = std::tie(&SomeClass::v1, &SomeClass::v2, &SomeClass::v3);
                                                                 ^
In file included from test.cpp:2:0:
/usr/include/c++/4.8/tuple:1044:5: error: in passing argument 1 of ‘std::tuple<_Elements& ...> std::tie(_Elements& ...) [with _Elements = {int SomeClass::*, int SomeClass::*, int SomeClass::*}]’
     tie(_Elements&... __args) noexcept
     ^

template< class... Types >
tuple<Types&...> tie( Types&... args );

auto t = std::make_tuple(&SomeClass::v1, &SomeClass::v2, &SomeClass::v3);

template< class... Types >
tuple<VTypes...> make_tuple( Types&&... args );
                                  ^^