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std::ranges::lexicographical_compare

Defined in header <algorithm>
Call signature 
template< std::input_iterator I1, std::sentinel_for<I1> S1,
          std::input_iterator I2, std::sentinel_for<I2> S2,
          class Proj1 = std::identity, class Proj2 = std::identity,
          std::indirect_strict_weak_order<
              std::projected<I1, Proj1>,
              std::projected<I2, Proj2>> Comp = ranges::less >
constexpr bool
    lexicographical_compare( I1 first1, S1 last1, I2 first2, S2 last2,
                             Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );
 (1) (since C++20) 
template< ranges::input_range R1, ranges::input_range R2,
          class Proj1 = std::identity, class Proj2 = std::identity,
          std::indirect_strict_weak_order<
              std::projected<ranges::iterator_t<R1>, Proj1>,
              std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less >
constexpr bool
    lexicographical_compare( R1&& r1, R2&& r2, Comp comp = {},
                             Proj1 proj1 = {}, Proj2 proj2 = {} );
 (2) (since C++20)

Checks if the first range [first1, last1) is lexicographically less than the second range [first2, last2).

1) Elements are compared using the given binary comparison function comp.
2) Same as (1), but uses r as the source range, as if using ranges::begin(r) as first and ranges::end(r) as last.

Lexicographical comparison is an 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.

The function-like entities described on this page are niebloids, that is:

In practice, they may be implemented as function objects, or with special compiler extensions.

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 first range of elements
proj2 - projection to apply to the second range of elements

Return value

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

Complexity

At most 2·min(N1, N2) applications of the comparison and corresponding projections, where N1 = ranges::distance(first1, last1) and N2 = ranges::distance(first2, last2).

Possible implementation

struct lexicographical_compare_fn
{
    template<std::input_iterator I1, std::sentinel_for<I1> S1,
             std::input_iterator I2, std::sentinel_for<I2> S2,
             class Proj1 = std::identity, class Proj2 = std::identity,
             std::indirect_strict_weak_order<
                 std::projected<I1, Proj1>,
                 std::projected<I2, Proj2>> Comp = ranges::less>
    constexpr bool operator()(I1 first1, S1 last1, I2 first2, S2 last2,
                              Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        for (; (first1 != last1) && (first2 != last2); ++first1, (void) ++first2)
        {
            if (std::invoke(comp, std::invoke(proj1, *first1), std::invoke(proj2, *first2)))
                return true;
 
            if (std::invoke(comp, std::invoke(proj2, *first2), std::invoke(proj1, *first1)))
                return false;
        }
        return (first1 == last1) && (first2 != last2);
    }
 
    template<ranges::input_range R1, ranges::input_range R2,
             class Proj1 = std::identity, class Proj2 = std::identity,
             std::indirect_strict_weak_order<
                 std::projected<ranges::iterator_t<R1>, Proj1>,
                 std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less>
    constexpr bool operator()(R1&& r1, R2&& r2, Comp comp = {},
                              Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        return (*this)(ranges::begin(r1), ranges::end(r1),
                       ranges::begin(r2), ranges::end(r2),
                       std::ref(comp), std::ref(proj1), std::ref(proj2));
    }
};
 
inline constexpr lexicographical_compare_fn lexicographical_compare;

Example

#include <algorithm>
#include <iostream>
#include <iterator>
#include <random>
#include <vector>
 
int main()
{
    std::vector<char> v1 {'a', 'b', 'c', 'd'};
    std::vector<char> v2 {'a', 'b', 'c', 'd'};
 
    namespace ranges = std::ranges;
    auto os = std::ostream_iterator<char>(std::cout, " ");
 
    std::mt19937 g {std::random_device {}()};
    while (not ranges::lexicographical_compare(v1, v2))
    {
        ranges::copy(v1, os);
        std::cout << ">= ";
        ranges::copy(v2, os);
        std::cout << '\n';
 
        ranges::shuffle(v1, g);
        ranges::shuffle(v2, g);
    }
 
    ranges::copy(v1, os);
    std::cout << "<  ";
    ranges::copy(v2, os);
    std::cout << '\n';
}

Possible output:

a b c d >= a b c d
d a b c >= c b d a
b d a c >= a d c b
a c d b <  c d a b

See also

(C++20)
determines if two sets of elements are the same
(niebloid)
returns true if one range is lexicographically less than another
(function template)

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