std::forward() 的右值引用重载的目的是什么？
What is the purpose of std::forward()'s rvalue reference overload?
我正在尝试完美转发，我发现
std::forward() 需要两个重载：
过载编号。 1:
template <typename T>
inline T&& forward(typename
std::remove_reference<T>::type& t) noexcept
{
return static_cast<T&&>(t);
}
2 号过载：
template <typename T>
inline T&& forward(typename
std::remove_reference<T>::type&& t) noexcept
{
static_assert(!std::is_lvalue_reference<T>::value,
"Can not forward an rvalue as an lvalue.");
return static_cast<T&&>(t);
}
现在完美转发的典型场景是
template <typename T>
void wrapper(T&& e)
{
wrapped(forward<T>(e));
}
当然你知道wrapper()实例化的时候，T取决于传递给它的参数是左值还是右值。如果它是 U 类型的左值，T 将推导为 U&。如果它是一个右值，T 被推断为 U.
在任何情况下 - 在 wrapper() 范围内 - e 是一个左值，因此它总是使用 std::forward().
的第一个重载
现在我的问题是：
使用（并且需要）第二次重载的有效场景是什么？
forward 的设计原理在 N2951 中进行了详细讨论。
本文档列出了 6 个用例：
A. Should forward an lvalue as an lvalue. All implementations pass
this test. But this is not the classic perfect forwarding pattern. The
purpose of this test is to show that implementation 2 fails in its
stated goal of preventing all use cases except perfect forwarding.
B. Should forward an rvalue as an rvalue. Like use case A, this is
an identity transformation and this presents a motivating example
where the identity transformation is needed.
C. Should not forward an rvalue as an lvalue. This use case
demonstrates a dangerous situation of accidentally creating a dangling
reference.
D. Should forward less cv-qualified expressions to more
cv-qualified expressions. A motivating use case involving the
addition of const during the forward.
E. Should forward expressions of derived type to an accessible,
unambiguous base type. A motivating use case involving forwarding a
derived type to a base type.
F. Should not forward arbitrary type conversions. This use case
demonstrates how arbitrary conversions within a forward lead to
dangling reference run time errors.
第二个重载启用案例 B 和 C。
本文继续提供每个用例的示例，这些示例太长，无法在此处重复。
更新
我只是 运行 第一个重载的 "solution" 通过这 6 个用例，这个练习表明第二个重载也启用用例 F：不应该转发任意类型转换。
我正在尝试完美转发，我发现
std::forward() 需要两个重载：
过载编号。 1:
template <typename T>
inline T&& forward(typename
std::remove_reference<T>::type& t) noexcept
{
return static_cast<T&&>(t);
}
2 号过载：
template <typename T>
inline T&& forward(typename
std::remove_reference<T>::type&& t) noexcept
{
static_assert(!std::is_lvalue_reference<T>::value,
"Can not forward an rvalue as an lvalue.");
return static_cast<T&&>(t);
}
现在完美转发的典型场景是
template <typename T>
void wrapper(T&& e)
{
wrapped(forward<T>(e));
}
当然你知道wrapper()实例化的时候，T取决于传递给它的参数是左值还是右值。如果它是 U 类型的左值，T 将推导为 U&。如果它是一个右值，T 被推断为 U.
在任何情况下 - 在 wrapper() 范围内 - e 是一个左值，因此它总是使用 std::forward().
现在我的问题是：
使用（并且需要）第二次重载的有效场景是什么？
forward 的设计原理在 N2951 中进行了详细讨论。
本文档列出了 6 个用例：
A. Should forward an lvalue as an lvalue. All implementations pass this test. But this is not the classic perfect forwarding pattern. The purpose of this test is to show that implementation 2 fails in its stated goal of preventing all use cases except perfect forwarding.
B. Should forward an rvalue as an rvalue. Like use case A, this is an identity transformation and this presents a motivating example where the identity transformation is needed.
C. Should not forward an rvalue as an lvalue. This use case demonstrates a dangerous situation of accidentally creating a dangling reference.
D. Should forward less cv-qualified expressions to more cv-qualified expressions. A motivating use case involving the addition of const during the forward.
E. Should forward expressions of derived type to an accessible, unambiguous base type. A motivating use case involving forwarding a derived type to a base type.
F. Should not forward arbitrary type conversions. This use case demonstrates how arbitrary conversions within a forward lead to dangling reference run time errors.
第二个重载启用案例 B 和 C。
本文继续提供每个用例的示例，这些示例太长，无法在此处重复。
更新
我只是 运行 第一个重载的 "solution" 通过这 6 个用例，这个练习表明第二个重载也启用用例 F：不应该转发任意类型转换。