auto、decltype(auto) 和尾随 return 类型

Question

有区别吗:

template <class T>
constexpr decltype(auto) f(T&& x) -> decltype(std::get<0>(std::forward<T>(x)))
{
    return std::get<0>(std::forward<T>(x));
}

和：

template <class T>
constexpr auto f(T&& x) -> decltype(std::get<0>(std::forward<T>(x)))
{
    return std::get<0>(std::forward<T>(x));
}

如果是的话，它是什么，我应该使用哪个来完美转发？

Answer 1

尾随 return 类型只能与 auto

一起使用

decltype(auto) vs auto 的重点是distinguish the case whether the return type should be a reference or value。但是在您的情况下，return 类型已经明确定义为 decltype(std::get<0>(std::forward<T>(x)))，因此即使您使用 auto.

，它也会被完美转发

在auto f() -> T中，"auto"关键字只是一个syntactic construct to fill in a type position。它没有其他用途。

---

事实上，在 C++17 中，您 不能 将 decltype(auto) 与尾随 return 类型一起使用。

C++14 措辞 (n3936 §7.1.6.4[dcl.spec.auto]/1):

The auto and decltype(auto) type-specifiers designate a placeholder type that will be replaced later, either by deduction from an initializer or by explicit specification with a trailing-return-type. The auto type-specifier is also used to signify that a lambda is a generic lambda.

C++17 措辞 (n4618 §7.1.7.4[dcl.spec.auto]/1):

The auto and decltype(auto) type-specifiers are used to designate a placeholder type that will be replaced later by deduction from an initializer. The auto type-specifier is also used to introduce a function type having a trailing-return-type or to signify that a lambda is a generic lambda (5.1.5). The auto type-specifier is also used to introduce a decomposition declaration (8.5).

这是DR 1852, see Does a placeholder in a trailing-return-type override an initial placeholder?。

实际上，虽然 gcc 接受 decltype(auto) f() -> T (which is a bug)，但 clang 会 拒绝 它说

error: function with trailing return type must specify return type 'auto',
not 'decltype(auto)'