< cpp‎ | container‎ | list

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

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 this == &other. 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.


[edit] Parameters

other - another container to merge
comp - 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);

The signature does not need to have const &, but the function object must not modify the objects passed to it.
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. ​

[edit] Return value


[edit] Example

#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 };
    std::cout << "list1:  " << list1 << "\n";
    std::cout << "list2:  " << list2 << "\n";
    std::cout << "merged: " << list1 << "\n";


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

[edit] Complexity

at most size() + other.size() - 1 comparisons.

[edit] See also

moves elements from another list
(public member function)