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std::ranges...cal_compare(3) C++ Standard Libary std::ranges...cal_compare(3) NAME std::ranges::lexicographical_compare - std::ranges::lexicographi- cal_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>, std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less (2) (since C++20) > 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 a operation with the following proper- ties: * Two ranges are compared element by element. * The first mismatching element defines which range is lexico- graphically less or greater than the other. * If one range is a prefix of another, the shorter range is lexi- cographically 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 may not be specified when call- ing any of them. * None of them is visible to argument-dependent lookup. * When one of them is found by normal unqualified lookup for the name to the left of the function-call operator, it inhibits argument-dependent lookup. 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 ele- ments 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 2min(N1, N2) applications of the comparison and correspond- ing 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 <iterator> #include <iostream> #include <vector> #include <random> 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 (!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 lexicographi- cally less than lexicographical_compare another (function template) http://cppreference.com 2022.07.31 std::ranges...cal_compare(3)
NAME | Synopsis | Parameters | Return value | Complexity | Possible implementation | Example | Possible output: | See also
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