std::list<T,Allocator>::merge (3) - Linux Manuals

std::list<T,Allocator>::merge: std::list<T,Allocator>::merge

NAME

std::list<T,Allocator>::merge - std::list<T,Allocator>::merge

Synopsis

void merge( list& other ); (1)
void merge( list&& other ); (1) (since C++11)
template <class Compare> (2)
void merge( list& other, Compare comp );
template <class Compare> (2) (since C++11)
void merge( list&& other, Compare comp );

Merges two sorted lists into one. The lists should be sorted into ascending order.
No elements are copied. The container other becomes empty after the operation. The function does nothing if other refers to the same object as *this. If get_allocator() != other.get_allocator(), the behavior is undefined. No iterators or references become invalidated, except that the iterators of moved elements now refer into *this, not into other. The first version uses operator< to compare the elements, the second version uses the given comparison function comp.
This operation is stable: for equivalent elements in the two lists, the elements from *this shall always precede the elements from other, and the order of equivalent elements of *this and other does not change.

Parameters

other - another container to merge
        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 an object of type list<T,Allocator>::const_iterator can be dereferenced and then implicitly converted to both of them.

Return value

(none)

Exceptions

If an exception is thrown, this function has no effect (strong exception guarantee), except if the exception comes from the comparison function.

Complexity

at most std::distance(begin(), end()) + std::distance(other.begin(), other.end()) - 1 comparisons.

Example

// Run this code

  #include <iostream>
  #include <list>

  std::ostream& operator<<(std::ostream& ostr, const std::list<int>& list)
  {
      for (auto &i : list) {
          ostr << " " << i;
      }
      return ostr;
  }

  int main()
  {
      std::list<int> list1 = { 5,9,0,1,3 };
      std::list<int> list2 = { 8,7,2,6,4 };

      list1.sort();
      list2.sort();
      std::cout << "list1: " << list1 << "\n";
      std::cout << "list2: " << list2 << "\n";
      list1.merge(list2);
      std::cout << "merged: " << list1 << "\n";
  }

Output:

  list1: 0 1 3 5 9
  list2: 2 4 6 7 8
  merged: 0 1 2 3 4 5 6 7 8 9