[stmt.expand] iter initialization for iterating expansion statements

[jakubjelinek]

Full name of submitter (unless configured in github; will be published with the issue): Jakub Jelinek

Reference (section label): [stmt.expand]

Link to reflector thread (if any):

Issue description:
[stmt.expand]/(5.2) includes

  static constexpr auto iter = begin + i;

where clearly i is an integer constant going from 0 to N-1, but doesn't say what type i has. For enumerating and destructuring expansion statements I think that doesn't matter, but begin + i is often a call to an overloaded operator and in that case it matters if i
has say std::size_t type, or std::ptrdiff_t, or int, or unsigned long long. Depending on that the code could work differently.

Suggested resolution:

[jakubjelinek]

  static constexpr auto iter = begin + std::ptrdiff_t(i);

would make it clearer.

[Eisenwave]

Arbitrarily using std::ptrdiff_t is not a good idea; there's no guarantee that this is accepted as a difference type by iterators.

What we can do is say:

 and Si is
 {
   static constexpr auto iter = begin + i;
   for-range-declaration = *iter;
   compound-statement
 }
+where i is a prvalue of type iter_difference_t<decltype(begin)>.
 The variables range, begin, end, and iter are defined for exposition only.

[Eisenwave]

Actually, I'm wrong on that one. https://eel.is/c++draft/iterator.concept.winc#6 states that expressions of integer type are convertible to integer-like-class types, and for iterators, the difference_type has to satisfy that requirement. Therefore, using std::ptrdiff_t would work, but I wouldn't want to conjure i out of thin air and wrap it in std::ptrdiff_t(i). It's still unclear what such an expression would mean.