c++ - auto、decltype(auto) 和尾随返回类型

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，它是什么，我应该使用哪个来完美转发？

Answer 1

尾随返回类型只能与 auto 一起使用

点decltype(auto)对比auto是distinguish the case whether the return type should be a reference or value .但在您的情况下，返回类型已明确定义为 decltype(std::get<0>(std::forward<T>(x))) , 因此即使您使用 auto 也会被完美转发.

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

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事实上，在 C++17 中你不能使用 decltype(auto)与 trailing-return-type 一起。

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 , 请参阅 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 )，但是 clang会拒绝它说

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