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operator==,!=(std::unordered_multiset)

From cppreference.com
 
 
 
 
template< class Key, class Hash, class KeyEqual, class Allocator >

bool operator==( const unordered_multiset<Key,Hash,KeyEqual,Allocator>& lhs,

                 const unordered_multiset<Key,Hash,KeyEqual,Allocator>& rhs );
(1)
template< class Key, class Hash, class KeyEqual, class Allocator >

bool operator!=( const unordered_multiset<Key,Hash,KeyEqual,Allocator>& lhs,

                 const unordered_multiset<Key,Hash,KeyEqual,Allocator>& rhs );
(2)

Compares the contents of two unordered containers.

The contents of two unordered containers lhs and rhs are equal if the following conditions hold:

  • lhs.size() == rhs.size()
  • each group of equivalent elements [lhs_eq1, lhs_eq2) obtained from lhs.equal_range(lhs_eq1) has a corresponding group of equivalent elements in the other container [rhs_eq1, rhs_eq2) obtained from rhs.equal_range(rhs_eq1), that has the following properties:

The behavior is undefined if Key is not EqualityComparable.

The behavior is also undefined if Hash and KeyEqual do not have the same behavior on lhs and rhs or if operator== for value_type is not a refinement of the partition into equivalent-key groups introduced by KeyEqual (that is, if two elements that compare equal using operator== fall into different partitions)

[edit] Parameters

lhs, rhs - unordered containers to compare

[edit] Return value

1) true if the contents of the containers are equal, false otherwise
2) true if the contents of the containers are not equal, false otherwise

[edit] Complexity

Proportional to ΣSi2 calls to operator== on value_type, calls to the predicate returned by key_eq, and calls to the hasher returned by hash_function in the average case, where S is the size of the ith equivalent key group. Proprotional to N2 in the worst case, where N is the size of the container. Average case becomes proportional to N if the elements within each equivalent key group are arranged in the same order (happens when the containers are copies of each other).