27 Algorithms library [algorithms]

27.6 Non-modifying sequence operations [alg.nonmodifying]

27.6.8 Find end [alg.find.end]

template<class ForwardIterator1, class ForwardIterator2> constexpr ForwardIterator1 find_end(ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2); template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2> ForwardIterator1 find_end(ExecutionPolicy&& exec, ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2); template<class ForwardIterator1, class ForwardIterator2, class BinaryPredicate> constexpr ForwardIterator1 find_end(ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2, BinaryPredicate pred); template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2, class BinaryPredicate> ForwardIterator1 find_end(ExecutionPolicy&& exec, ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2, BinaryPredicate pred); template<forward_iterator I1, sentinel_for<I1> S1, forward_iterator I2, sentinel_for<I2> S2, class Pred = ranges::equal_to, class Proj1 = identity, class Proj2 = identity> requires indirectly_comparable<I1, I2, Pred, Proj1, Proj2> constexpr subrange<I1> ranges::find_end(I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {}); template<forward_range R1, forward_range R2, class Pred = ranges::equal_to, class Proj1 = identity, class Proj2 = identity> requires indirectly_comparable<iterator_t<R1>, iterator_t<R2>, Pred, Proj1, Proj2> constexpr borrowed_subrange_t<R1> ranges::find_end(R1&& r1, R2&& r2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {});
Let:
  • pred be equal_to{} for the overloads with no parameter pred;
  • E be:
    • pred(*(i + n), *(first2 + n)) for the overloads in namespace std;
    • invoke(pred, invoke(proj1, *(i + n)), invoke(proj2, *(first2 + n))) for the overloads in namespace ranges;
  • i be last1 if [first2, last2) is empty, or if (last2 - first2) > (last1 - first1) is true, or if there is no iterator in the range [first1, last1 - (last2 - first2)) such that for every non-negative integer n < (last2 - first2), E is true.
    Otherwise i is the last such iterator in [first1, last1 - (last2 - first2)).
Returns:
  • i for the overloads in namespace std.
  • {i, i + (i == last1 ? 0 : last2 - first2)} for the overloads in namespace ranges.
Complexity: At most (last2 - first2) * (last1 - first1 - (last2 - first2) + 1) applications of the corresponding predicate and any projections.