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std::ranges...cal_compare(3)   C++ Standard Libary  std::ranges...cal_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::be-
     gin(r) as
	first and ranges::end(r) as last.

	Lexicographical  comparison  is  an operation with the following proper-
     ties:

	  * Two ranges are compared element by element.
	  * The first mismatching element defines which range  is  lexicographi-
     cally less or
	    greater than the other.
	  *  If  one  range is a prefix of another, the shorter range is lexico-
     graphically 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  inhib-
     ited.

	In  practice,  they may be implemented as function objects, or with spe-
     cial 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 2A.min(N1, N2) applications of the comparison and  corresponding
     projections,
	where  N1  =  ranges::distance(first1,	last1)	and  N2  =  ranges::dis-
     tance(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::in-
     voke(proj2, *first2)))
		     return true;

		 if  (std::invoke(comp,  std::invoke(proj2,  *first2),	std::in-
     voke(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)

http://cppreference.com 	   2024.06.10	    std::ranges...cal_compare(3)
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