std::forward_list<T,Allocator>::merge
void merge( forward_list& other ); |
(1) | (since C++11) |
void merge( forward_list&& other ); |
(2) | (since C++11) |
template< class Compare > void merge( forward_list& other, Compare comp ); |
(3) | (since C++11) |
template< class Compare > void merge( forward_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.
If *this or other is not sorted with respected to the corresponding comparator, or get_allocator() != other.get_allocator(), the behavior is undefined.
Contents |
[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) |
Type requirements | ||
-Compare must meet the requirements of Compare.
|
[edit] Return value
(none)
[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()):
[edit] Example
#include <iostream> #include <forward_list> std::ostream& operator<<(std::ostream& ostr, const std::forward_list<int>& list) { for (const int i : list) ostr << ' ' << i; return ostr; } int main() { std::forward_list<int> list1 = {5, 9, 1, 3, 3}; std::forward_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
[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 2045 | C++11 | O(1) node moving could not be guaranteed if get_allocator() != other.get_allocator() |
the behavior is undefined in this case |
LWG 3088 | C++11 | the effect when *this and other refer to the same object was not specified operator< could misbehave for pointer elements |
specified as no-op implementation-defined strict total order used |
[edit] See also
moves elements from another forward_list (public member function) | |
merges two sorted ranges (function template) | |
merges two ordered ranges in-place (function template) | |
(C++20) |
merges two sorted ranges (algorithm function object) |
(C++20) |
merges two ordered ranges in-place (algorithm function object) |