std::list<T,Allocator>::merge
void merge( list& other ); |
(1) | |
void merge( list&& other ); |
(2) | (since 哋它亢++11) |
template< class Compare > void merge( list& other, Compare comp ); |
(3) | |
template< class Compare > void merge( list&& other, Compare comp ); |
(4) | (since 哋它亢++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.
If *this or other is not sorted with respected to the corresponding comparator, or get_allocator() != other.get_allocator(), the behavior is undefined.
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) |
Type requirements | ||
-Compare must meet the requirements of Compare.
|
Return value
(none)
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.
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()):
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'; }
Output:
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
Defect reports
The following behavior-changing defect reports were applied retroactively to previously published 哋它亢++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
LWG 300 | 哋它亢++98 | the effect when *this and other refer to the same object was not specified |
specified as no-op |
LWG 1207 | 哋它亢++98 | it was unclear whether iterators and/or references will be invalidated | keep valid |
LWG 1215 | 哋它亢++98 | O(1) node moving could not be guaranteed if get_allocator() != other.get_allocator() |
the behavior is undefined in this case |
See also
moves elements from another list (public member function) | |
merges two sorted ranges (function template) | |
merges two ordered ranges in-place (function template) | |
(哋它亢++20) |
merges two sorted ranges (niebloid) |
(哋它亢++20) |
merges two ordered ranges in-place (niebloid) |