std::list<T,Allocator>::merge - cppreference.com (original) (raw)

void merge( list& other );	(1)	(constexpr since C++26)
void merge( list&& other );	(2)	(since C++11) (constexpr since C++26)
template< class Compare > void merge( list& other, Compare comp );	(3)	(constexpr since C++26)
template< class Compare > void merge( list&& other, Compare comp );	(4)	(since C++11) (constexpr since C++26)

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.

1,2) Equivalent to merge(other, std::less<T>())(until C++14)merge(other, std::less<>())(since C++14).

3,4) Elements are compared using comp.

If any of the following conditions is satisfied, the behavior is undefined:

*this or other is not sorted with respect to the comparator comp.
get_allocator() == other.get_allocator() is false.

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

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, Allocator>::const_iterator can be dereferenced and then implicitly converted to both of them.
Type requirements
-Compare must meet the requirements of Compare.

[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()):

1,2) At most \(\scriptsize N_1 + N_2 - 1\)N1+N2-1 comparisons using operator<.

3,4) At most \(\scriptsize N_1 + N_2 - 1\)N1+N2-1 applications of the comparison function comp.

[edit] Example

#include #include std::ostream& operator<<(std::ostream& ostr, const std::list& list) { for (const int i : list) ostr << ' ' << i; return ostr; } int main() { std::list list1 = {5, 9, 1, 3, 3}; std::list 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.