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merge

merge PrototypeMerge is an overloaded name: there are actually two merge functions.
template <class InputIterator1, class InputIterator2, class OutputIterator>
OutputIterator merge(InputIterator1 first1, InputIterator1 last1,
                     InputIterator2 first2, InputIterator2 last2,
                     OutputIterator result);

template <class InputIterator1, class InputIterator2, class OutputIterator,
          class StrictWeakOrdering>
OutputIterator merge(InputIterator1 first1, InputIterator1 last1,
                     InputIterator2 first2, InputIterator2 last2,
                     OutputIterator result, StrictWeakOrdering comp);
DescriptionMerge combines two sorted ranges [first1, last1) and [first2, last2) into a single sorted range. That is, it copies elements from [first1, last1) and [first2, last2) into [result, result + (last1 - first1) + (last2 - first2)) such that the resulting range is in ascending order. Merge is stable, meaning both that the relative order of elements within each input range is preserved, and that for equivalent [1] elements in both input ranges the element from the first range precedes the element from the second. The return value is result + (last1 - first1) + (last2 - first2).

The two versions of merge differ in how elements are compared. The first version uses operator<. That is, the input ranges and the output range satisfy the condition that for every pair of iterators i and j such that i precedes j, *j < *i is false. The second version uses the function object comp. That is, the input ranges and the output range satisfy the condition that for every pair of iterators i and j such that i precedes j, comp(*j, *i) is false.

Definition Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h. Requirements on types For the first version: For the second version: Preconditions For the first version: For the second version: Complexity Linear. No comparisons if both [first1, last1) and [first2, last2) are empty ranges, otherwise at most (last1 - first1) + (last2 - first2) - 1 comparisons. Example
int main()
{
  int A1[] = { 1, 3, 5, 7 };
  int A2[] = { 2, 4, 6, 8 };
  const int N1 = sizeof(A1) / sizeof(int);
  const int N2 = sizeof(A2) / sizeof(int);

  merge(A1, A1 + N1, A2, A2 + N2, 
        ostream_iterator<int>(cout, " "));
  // The output is "1 2 3 4 5 6 7 8"
}
Notes

[1] Note that you may use an ordering that is a strict weak ordering but not a total ordering; that is, there might be values x and y such that x < y, x > y, and x == y are all false. (See the LessThan Comparable requirements for a more complete discussion.) Two elements x and y are equivalent if neither x < y nor y < x. If you're using a total ordering, however (if you're using strcmp, for example, or if you're using ordinary arithmetic comparison on integers), then you can ignore this technical distinction: for a total ordering, equality and equivalence are the same.

See alsoinplace_merge, set_union, sort STL Main Page

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