使用 std：：make_tuple 实例化具有显式指定类型的元组与使用元组构造函数 [duplicate] 之间的区别

问：

以下面的代码片段为例，该代码片段使用 std：：make_tuple 函数和元组构造函数实例化一些元组实例：

...

const auto integer = 0;
const auto text = std::string{};

/* tuple instance constructed using std::make_tuple with implicit type deduction */
const auto tuple_instance_with_implicit_make = std::make_tuple(integer, text);   // OK
    
/* tuple instance constructed using std::make_tuple with explicitly specified types */
const auto tuple_instance_with_explicit_make = std::make_tuple<int, std::string>(integer, text); // ERROR
    
/* tuple instance using tuple constructor with explicitly specified types */
const auto tuple_instance_explicit = std::tuple<int, std::string>(integer, text); // OK

/* tuple instance using tuple constructor with implicit type deduction */
const auto tuple_instance_implicit = std::tuple(integer, text);                   // OK
    
...

当我运行代码时，对于使用具有显式指定类型的 std：：make_tuple 函数的情况，我收到以下错误，但对于直接将元组构造函数与显式指定类型一起使用的情况，则不会出现以下错误：

 error: cannot bind rvalue reference of type 'int&&' to lvalue of type 'const int'
   12 |     const auto tuple_instance_with_explicit_make = std::make_tuple<int, std::string>(integer, text); // ERROR
      |                                                                                      ^~~~~~~
In file included from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/bits/uses_allocator_args.h:38,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/bits/memory_resource.h:41,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/string:58,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/bits/locale_classes.h:40,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/bits/ios_base.h:41,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/ios:44,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/ostream:40,
                 from /opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/iostream:41,
                 from <source>:1:
/opt/compiler-explorer/gcc-13.2.0/include/c++/13.2.0/tuple:1987:27: note:   initializing argument 1 of 'constexpr std::tuple<typename std::__strip_reference_wrapper<typename std::decay<_Elements>::type>::__type ...> std::make_tuple(_Elements&& ...) [with _Elements = {int, __cxx11::basic_string<char, char_traits<char>, allocator<char> >}]'
 1987 |     make_tuple(_Elements&&... __args)
      |                ~~~~~~~~~~~^~~~~~~~~~

由于 std：：make_tuple 函数和元组构造函数都对其参数使用转发引用，并且 std：：make_tuple 将传递的参数转发给元组构造函数，因此我本来希望具有相同的行为。据我所知，转发引用可以绑定到任何值类别的对象和表达式，无论是左值还是右值。

有人可以帮我理解这种行为吗？为什么使用具有显式指定类型的构造函数创建元组实例有效，但以相同方式使用 std：：make_tuple 的元组实例却不行？