std::ranges::lexicographical_compare(3)

NAME
std::ranges::lexicographical_compare - std::ranges::lexicographical_compare

Synopsis
   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>, (1) (since C++20)
                 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 = {} );
   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>, (2) (since C++20)
                 std::projected<ranges::iterator_t<R2>, Proj2>> Comp
   = ranges::less >
   constexpr bool
       lexicographical_compare( R1&& r1, R2&& r2, Comp comp = {},
                        Proj1 proj1 = {}, Proj2 proj2 = {} );
   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:
     * Explicit template argument lists cannot be specified when calling any of them.
     * None of them are visible to argument-dependent lookup.
     * When any of them are found by normal unqualified lookup as the name to the left
       of the function-call operator, argument-dependent lookup is inhibited.
   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
// Run this code
 #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
   ranges::equal determines if two sets of elements are the same
   (C++20) (niebloid)
                        returns true if one range is lexicographically less than
   lexicographical_compare another
                        (function template)