Created on 2019-10-27.00:00:00 last changed 7 months ago
This wording is relative to N4835.
Modify [range.iter.op.advance] as indicated:
template<input_or_output_iterator I, sentinel_for<I> S> constexpr void ranges::advance(I& i, S bound);
-3- Expects: [i, bound) denotes a range.
(4.1) — If I and S model assignable_from<I&, S>, equivalent to i = std::move(bound).
(4.2) — Otherwise, if S and I model sized_sentinel_for<S, I>, equivalent to ranges::advance(i, bound - i).
(4.3) — Otherwise, while bool(i != bound) is true, increments i.
[ 2019-11 Priority to 2 during Monday issue prioritization in Belfast ]
Recall that "[i, s) denotes a range" for an iterator i and sentinel s means that either i == s holds, or i is dereferenceable and [++i, s) denotes a range ([iterator.requirements.general).
The three-argument overload ranges::advance(i, n, bound) is specified in [range.iter.op.advance] paragraphs 5 through 7. Para 5 establishes a precondition that [bound, i) denotes a range when n < 0 (both bound and i must have the same type in this case). When sized_sentinel_for<S, I> holds and n < bound - i, para 6.1.1 says that ranges::advance(i, n, bound) is equivalent to ranges::advance(i, bound). Para 3, however, establishes a precondition for ranges::advance(i, bound) that [i, bound) denotes a range. [bound, i) and [i, bound) cannot both denote ranges unless i == bound, which is not the case for all calls that reach 6.1.1.
The call in para 6.1.1 wants the effects of either 4.1 - which really has no preconditions - or 4.2, which is well-defined if either [i, bound) or [bound, i) denotes a range. Para 3's stronger precondition is actually only required by Para 4.3, which increments i blindly looking for bound. The straight-forward fix here seems to be to relax para 3's precondition to only apply when 4.3 will be reached.
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