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std::merge(3) C++ Standard Libary std::merge(3) NAME std::merge - std::merge Synopsis Defined in header <algorithm> template< class InputIt1, class InputIt2, class OutputIt > OutputIt merge( InputIt1 first1, InputIt1 last1, (1) (const- expr since C++20) InputIt2 first2, InputIt2 last2, OutputIt d_first ); template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2, class ForwardIt3 > ForwardIt3 merge( ExecutionPolicy&& policy, (2) (since C++17) ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, ForwardIt3 d_first ); template< class InputIt1, class InputIt2, class OutputIt, class Compare > OutputIt merge( InputIt1 first1, InputIt1 last1, (3) (const- expr since C++20) InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp ); template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2, class ForwardIt3, class Compare > ForwardIt3 merge( ExecutionPolicy&& policy, (4) (since C++17) ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, ForwardIt3 d_first, Compare comp ); Merges two sorted ranges [first1, last1) and [first2, last2) into one sorted range beginning at d_first. 1) If [first1, last1) or [first2, last2) is not sorted with respect to operator< (until C++20) std::less{} (since C++20), the behavior is undefined. 3) If [first1, last1) or [first2, last2) is not sorted with respect to comp, the behavior is undefined. 2,4) Same as (1,3), but executed according to policy. These overloads participate in overload resolution only if std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> is true. (until C++20) std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>> is true. (since C++20) This merge function is stable, which means that for equivalent ele- ments in the original two ranges, the elements from the first range (preserving their original order) precede the elements from the second range (preserving their original order). If the output range overlaps with [first1, last1) or [first2, last2), the behavior is undefined. Parameters first1, last1 - the first range of elements to merge first2, last2 - the second range of elements to merge d_first - the beginning of the destination range policy - the execution policy to use. See execution policy for details. comparison function object (i.e. an object that sat- isfies the requirements of Compare) which returns true if the first argument is less than (i.e. is ordered before) the second. The signature of the comparison function should be equivalent to the following: bool cmp(const Type1& a, const Type2& b); comp - While the signature does not need to have const&, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) Type1 and Type2 re- gardless of value category (thus, Type1& is not allowed , nor is Type1 unless for Type1 a move is equivalent to a copy (since C++11)). The types Type1 and Type2 must be such that objects of types InputIt1 and InputIt2 can be dereferenced and then implicitly converted to both Type1 and Type2. Type requirements - InputIt1, InputIt2 must meet the requirements of LegacyInputItera- tor. - ForwardIt1, ForwardIt2, ForwardIt3 must meet the requirements of LegacyForwardIterator. - OutputIt must meet the requirements of LegacyOutputIterator. - Compare must meet the requirements of Compare. Return value An output iterator to element past the last element copied. Complexity Given \(\scriptsize N_1\)N 1 as std::distance(first1, last1) and \(\scriptsize N_2\)N 2 as std::distance(first2, last2): 1) At most \(\scriptsize N_1+N_2-1\)N 1+N 2-1 comparisons using operator< (until C++20) std::less{} (since C++20). 2) \(\scriptsize O(N_1+N_2)\)O(N 1+N 2) comparisons using operator< (until C++20) std::less{} (since C++20). 3) At most \(\scriptsize N_1+N_2-1\)N 1+N 2-1 applications of the comparison function comp. 4) \(\scriptsize O(N_1+N_2)\)O(N 1+N 2) applications of the comparison function comp. Exceptions The overloads with a template parameter named ExecutionPolicy report errors as follows: * If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the standard policies, std::termi- nate is called. For any other ExecutionPolicy, the behavior is implementation- defined. * If the algorithm fails to allocate memory, std::bad_alloc is thrown. Possible implementation See also the implementations in libstdc++ and libc++. merge (1) template<class InputIt1, class InputIt2, class OutputIt> OutputIt merge(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first) { for (; first1 != last1; ++d_first) { if (first2 == last2) return std::copy(first1, last1, d_first); if (*first2 < *first1) { *d_first = *first2; ++first2; } else { *d_first = *first1; ++first1; } } return std::copy(first2, last2, d_first); } merge (3) template<class InputIt1, class InputIt2, class OutputIt, class Compare> OutputIt merge(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp) { for (; first1 != last1; ++d_first) { if (first2 == last2) return std::copy(first1, last1, d_first); if (comp(*first2, *first1)) { *d_first = *first2; ++first2; } else { *d_first = *first1; ++first1; } } return std::copy(first2, last2, d_first); } Notes This algorithm performs a similar task as std::set_union does. Both consume two sorted input ranges and produce a sorted output with elements from both inputs. The difference between these two algorithms is with handling values from both input ranges which compare equivalent (see notes on LessThanComparable). If any equivalent values appeared n times in the first range and m times in the sec- ond, std::merge would output all n + m occurrences whereas std::set_union would out- put std::max(n, m) ones only. So std::merge outputs exactly std::distance(first1, last1) + std::distance(first2, last2) values and std::set_union may produce fewer. Example // Run this code #include <algorithm> #include <functional> #include <iostream> #include <iterator> #include <random> #include <vector> auto print = [](const auto rem, const auto& v) { std::cout << rem; std::copy(v.begin(), v.end(), std::ostream_itera- tor<int>(std::cout, " ")); std::cout << '\n'; }; int main() { // fill the vectors with random numbers std::random_device rd; std::mt19937 mt(rd()); std::uniform_int_distribution<> dis(0, 9); std::vector<int> v1(10), v2(10); std::generate(v1.begin(), v1.end(), std::bind(dis, std::ref(mt))); std::generate(v2.begin(), v2.end(), std::bind(dis, std::ref(mt))); print("Originally:\nv1: ", v1); print("v2: ", v2); std::sort(v1.begin(), v1.end()); std::sort(v2.begin(), v2.end()); print("After sorting:\nv1: ", v1); print("v2: ", v2); // merge std::vector<int> dst; std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(dst)); print("After merging:\ndst: ", dst); } Possible output: Originally: v1: 2 6 5 7 4 2 2 6 7 0 v2: 8 3 2 5 0 1 9 6 5 0 After sorting: v1: 0 2 2 2 4 5 6 6 7 7 v2: 0 0 1 2 3 5 5 6 8 9 After merging: dst: 0 0 0 1 2 2 2 2 3 4 5 5 5 6 6 6 7 7 8 9 Defect reports The following behavior-changing defect reports were applied retroac- tively to previously published C++ standards. DR Applied to Behavior as published Correct be- havior LWG 780 C++98 the merge operation was not defined defined See also inplace_merge merges two ordered ranges in-place (function template) is_sorted checks whether a range is sorted into ascending order (C++11) (function template) set_union computes the union of two sets (function template) sort sorts a range into ascending order (function template) sorts a range of elements while preserving order be- tween equal stable_sort elements (function template) ranges::merge merges two sorted ranges (C++20) (niebloid) http://cppreference.com 2024.06.10 std::merge(3)
NAME | Synopsis | Parameters | Type requirements | Return value | Complexity | Exceptions | Possible implementation | Notes | Example | Possible output: | See also
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