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

The function does nothing if other refers to the same object as *this.

Otherwise, merges other into *this. Both lists should be sorted. No elements are copied, and the container other becomes 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, and 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.

1,2) Elements are compared using std::less<T>()(until C++14)std::less<>()(since C++14).
3,4) Elements are compared using comp.

If *this or other is not sorted with respected to the corresponding comparator, or get_allocator() != other.get_allocator(), the behavior is undefined.


[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);

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,&nbsp;Allocator>::const_iterator can be dereferenced and then implicitly converted to both of them. ​

Type requirements
Compare must meet the requirements of Compare.

[edit] Return value


[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 N as std::distance(begin(), end()) and R as std::distance(other.begin(), other.end()):

1,2) At most N+R-1 comparisons using operator<.
3,4) At most N+R-1 applications of the comparison function comp.

[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};
    std::cout << "list1: " << list1 << '\n';
    std::cout << "list2: " << list2 << '\n';
    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.

DR Applied to Behavior as published Correct behavior
LWG 300 C++98 the effect when *this and other refer
to the same object was not specified
specified as no-op
LWG 1207 C++98 it was unclear whether iterators and/or references will be invalidated keep valid
LWG 1215 C++98 O(1) node moving could not be guaranteed if
get_allocator() != other.get_allocator()
the behavior is
undefined in this case
LWG 3088 C++98 operator< could misbehave for pointer elements implementation-defined
strict total order used

[edit] See also

moves elements from another list
(public member function) [edit]
merges two sorted ranges
(function template) [edit]
merges two ordered ranges in-place
(function template) [edit]
merges two sorted ranges
merges two ordered ranges in-place