c++ decltype(auto) 或 decltype(std::forward<T>(value))？

Question

例如，简单的恒等函子：

template <typename T>
class identity
{
public:
    constexpr auto operator ()(T && i) -> decltype(std::forward<T>(i))
    {
        return std::forward<T>(i);
    }
};

return 值更好（C++14 和更新版本）：

-> decltype(std::forward<T>(i)) 或
-> decltype(auto)

还是它们相同？

Answer 1

Or are they the same?

假设你写对了：

constexpr decltype(auto) operator ()(T && i)
{
    return std::forward<T>(i);
}

他们是一样的。 [dcl.type.auto.deduct]:

A type T containing a placeholder type, and a corresponding initializer e, are determined as follows:

for a non-discarded return statement that occurs in a function declared with a return type that contains a placeholder type, T is the declared return type and e is the operand of the return statement. If the return statement has no operand, then e is void();

If the placeholder is the decltype(auto) type-specifier, T shall be the placeholder alone. The type deduced for T is determined as described in [dcl.type.simple], as though e had been the operand of the decltype

函数的 return 类型是从 return e; 推导出来的，就像 decltype(e) 一样。所以它与显式 decltype(std::forward<T>(i)).

相同

What is better

在这种情况下，我会选择 "less is more"。 decltype(auto) 以更简洁的方式为您提供所需内容。

c++ decltype(auto) 或 decltype(std::forward<T>(value))？

c++ decltype(auto) or decltype(std::forward<T>(value))?

c++

decltype

auto

c++14