What is the difference between std::tie and std::forward_as_tuple
Let's just look at the signatures. std::tie()
is:
template< class... Types > constexpr tuple<Types&...> tie( Types&... args ) noexcept;
whereas std::forward_as_tuple()
is:
template< class... Types > constexpr tuple<Types&&...> forward_as_tuple( Types&&... args ) noexcept;
The only difference is that the former accepts only lvalues whereas the latter accepts lvalues and rvalues. If all of your inputs are lvalues, as they are in your use-case, they are exactly equivalent.
std::tie()
is largely intended as the left-hand side of assignment (e.g. std::tie(a, b) = foo;
to unpack a pair
), whereas std::forward_as_tuple()
is largely intended to pass things around in functions to avoid copies. But they can both be used to solve this problem. tie
is obviously quite a bit shorter, and arguably more well-known (the cppreference example for tie
uses it to implement operator<
), so that would get my vote.