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

Merges two sorted lists into one sorted list.

• If other refers to the same object as *this, does nothing.
• Otherwise, transfers all elements from other to *this. other is empty after the merge.

This operation is stable:

• For equivalent elements in the two lists, the elements from *this always precede the elements from other.
• The order of equivalent elements of *this and other does not change.

No iterators or references become invalidated. The pointers and references to the elements moved from *this, as well as the iterators referring to these elements, will refer to the same elements of *this, instead of other.

[edit] Parameters

[edit] Exceptions

If an exception is thrown for any reason, these functions have no effect (strong exception safety guarantee). Except if the exception comes from a comparison.

[edit] Complexity

If other refers to the same object as *this, no comparisons are performed.

Otherwise, given \(\scriptsize N_1\)N1 as std::distance(begin(), end()) and \(\scriptsize N_2\)N2 as std::distance(other.begin(), other.end()):

[edit] Example

#include <iostream>
#include <list>
 
std::ostream& operator<<(std::ostream& ostr, const std::list<int>& list)
{
    for (const int i : list)
        ostr << ' ' << i;
    return ostr;
}
 
int main()
{
    std::list<int> list1 = {5, 9, 1, 3, 3};
    std::list<int> list2 = {8, 7, 2, 3, 4, 4};
 
    list1.sort();
    list2.sort();
    std::cout << "list1: " << list1 << '\n';
    std::cout << "list2: " << list2 << '\n';
 
    list1.merge(list2);
    std::cout << "merged:" << list1 << '\n';
}

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

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[edit] See also