std::experimental::ranges::lexicographical_compare (3) - Linux Manuals

std::experimental::ranges::lexicographical_compare: std::experimental::ranges::lexicographical_compare

NAME

std::experimental::ranges::lexicographical_compare - std::experimental::ranges::lexicographical_compare

Synopsis

Defined in header <experimental/ranges/algorithm>
template< InputIterator I1, Sentinel<I1> S1, InputIterator I2, Sentinel<I2> S2,
class Proj1 = ranges::identity, class Proj2 = ranges::identity,
class Comp = ranges::less<> >
requires IndirectStrictWeakOrder<Comp, projected<I1, Proj1>, projected<I2, Proj2>> (1) (ranges TS)
bool lexicographical_compare(I1 first1, S1 last1, I2 first2, S2 last2,
Comp comp = Comp{},
Proj1 proj1 = Proj1{}, Proj2 proj2 = Proj2{});
template< InputRange R1, InputRange R2,
class Proj1 = ranges::identity, class Proj2 = ranges::identity,
class Comp = ranges::less<> >
requires IndirectStrictWeakOrder<Comp, projected<ranges::iterator_t<R1>, Proj1>, (2) (ranges TS)
projected<ranges::iterator_t<R2>, Proj2>>
bool lexicographical_compare(R1&& r1, R2&& r2, Comp comp = Comp{},
Proj1 proj1 = Proj1{}, Proj2 proj2 = Proj2{});

1) Checks if the first range [first1, last1) is lexicographically less than the second range [first2, last2). Elements are compared using the given binary comparison function comp, after being projected with proj1 and proj2 respectively.
2) Same as (1), but uses r1 as the first source range and r2 as the second source range, as if using ranges::begin(r1) as first1, ranges::end(r1) as last1, ranges::begin(r2) as first2, and ranges::end(r2) as last2.
Lexicographical comparison is a operation with the following properties:

* Two ranges are compared element by element.
* The first mismatching element defines which range is lexicographically less or greater than the other.
* If one range is a prefix of another, the shorter range is lexicographically less than the other.
* If two ranges have equivalent elements and are of the same length, then the ranges are lexicographically equal.
* An empty range is lexicographically less than any non-empty range.
* Two empty ranges are lexicographically equal.

Parameters

first1, last1 - the first range of elements to examine
r1 - the first range of elements to examine
first2, last2 - the second range of elements to examine
r2 - the second range of elements to examine
comp - comparison function to apply to the projected elements
proj1 - projection to apply to the elements in the first range
proj2 - projection to apply to the elements in the second range

Return value

true if the first range is lexicographically less than the second.

Complexity

At most 2·min(N1, N2) applications of the comparison operation, where N1 = last1 - first1 and N2 = last2 - first2.

Possible implementation

  template< InputIterator I1, Sentinel<I1> S1, InputIterator I2, Sentinel<I2> S2,
            class Proj1 = ranges::identity, class Proj2 = ranges::identity,
            class Comp = ranges::less<> >
    requires IndirectStrictWeakOrder<Comp, projected<I1, Proj1>, projected<I2, Proj2>>
  bool lexicographical_compare(I1 first1, S1 last1, I2 first2, S2 last2,
                               Comp comp = Comp{},
                               Proj1 proj1 = Proj1{}, Proj2 proj2 = Proj2{})
  {
      for ( ; (first1 != last1) && (first2 != last2); (void) ++first1, (void) ++first2 ) {
          if (ranges::invoke(comp, ranges::invoke(proj1, *first1),
                                   ranges::invoke(proj2, *first2))) return true;
          if (ranges::invoke(comp, ranges::invoke(proj2, *first2),
                                   ranges::invoke(proj1, *first1))) return false;
      }
      return (first1 == last1) && (first2 != last2);
  }

Example

 This section is incomplete
 Reason: no example