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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) (constexpr 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) (constexpr 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 elements 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 satisfies 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 regardless 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 LegacyInputIterator.
-
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::terminate 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 second, std::merge
would output all n + m occurrences whereas std::set_union would output 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_iterator<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 retroactively to
previously published C++ standards.


DR Applied to Behavior as published Correct behavior
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 between equal
stable_sort elements
(function template)
ranges::merge merges two sorted ranges
(C++20) (niebloid)

2024.06.10 http://cppreference.com