Proposed resolution:

This wording is relative to N4640.

1. Modify [tuple.creation] as indicated:

   template<class... TTypes>
     constexpr tuple<TTypes&...> tie(TTypes&... t) noexcept;
   

   -7- Returns: […]

   -?- [Note: The constructors and assignment operators provided by ignore shall be constexpr. — end note]

   -8- [Example: […] — end example]

[ 2024-07-25 Status changed: New → Resolved. ]

Resolved by p2968r2, approved in St. Louis.

[ 2017-03-04, Kona ]

Set priority to 3. P/R is incorrect; it should NOT be a note. Marshall to work with Eric to get better wording. STL says "use an exposition-only class".

The current PR for LWG 2773 changes std::ignore to be a constexpr variable. However it says nothing about whether using std::ignore in std::tie is a constant expression. I think the intent was clearly to allow this. Therefore I suggest we update the resolution to explicitly call this out in a note. (I don't think new normative wording is needed).

I propose we update the current PR as follows:

1. Keep the current changes proposed by the PR.

2. Add a note after [tuple.creation]/p7 (std::tie):

   [Note: The constructors and assignment operators provided by ignore shall be constexpr]

Perhaps LWG feels the existing wording is clear enough, but if not I think the above changes sufficiently clarify it.

The ability to constexpr assign to std::ignore can be important: Here is an extremely contrived example:

constexpr bool foo() {
  auto res = std::tie(std::ignore);
  std::get<0>(res) =42; 
  return true;
}
static_assert(foo());