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

Question

例如，简单的恒等仿函数:

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

对于返回值，什么更好(C++14 和更新版本):

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

或者它们是一样的吗？

Answer 1

Or are they the same?

假设你写得正确:

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

它们是一样的。 [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

函数的返回类型是从return e;推导出来的好像是 decltype(e) .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“少即是多”。 decltype(auto)以更简洁的方式为您提供所需内容。