Why does return std::forward<T> fail to perfect forward params typed w/ T&&?

Programming
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1

The P0849 paper adds the auto(x) expression to the language to enable perfect forwarding. This is needed because the decay_copy function, shown below, cannot perfect forward parameters declared with a universal/forwarding reference.

template<class T>
constexpr decay_t<T> decay_copy(T&& v) noexcept(
    is_nothrow_convertible_v<T, decay_t<T>>) {
    return std::forward<T>(v);
}

An issue arises when using decay_copy(x.front()): even if x.front() is a prvalue (a temporary object), it will be copied and not perfectly forwarded.

2

The issue arises because decay_copy(x.front()) is not able to perfectly forward a prvalue (temporary object) argument. Instead, it will be copied.