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unordered_multiset operator = in C++ STL

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  • Last Updated : 23 Jun, 2022
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The ‘=’ is an operator in C++ STL which copies (or moves) an unordered_multiset to another unordered_multiset and unordered_multiset::operator= is the corresponding operator function. There are three versions of this function:

  • The first version takes reference of an unordered_multiset as an argument and copies it to an unordered_multiset. Syntax:
ums1.operator=(unordered_multiset &ums2)
  • Parameters: The first version takes the reference of an unordered_multiset as argument.
  • The second version performs a move assignment i.e it moves the content of an unordered_multiset to another unordered_multiset. Syntax:
ums1.operator=(unordered_multiset &&ums2)
  • Parameters: The second version takes the r-value reference of an unordered_multiset as argument
  • The third version assigns contents of an initializer list to an unordered_multiset. Syntax:
ums1.operator=(initializer list)
  • Parameters: The third version takes an initializer list as argument.

Return Value: All of them return the value of this pointer(*this). The following program illustrates unordered_multiset::operator= . 

CPP




// C++ code to illustrate the method
// unordered_multiset::operator=()
 
#include <iostream>
#include <unordered_set>
using namespace std;
 
// merge function
template <class T>
 
T merge(T a, T b)
{
    T t(a);
    t.insert(b.begin(), b.end());
    return t;
}
 
int main()
{
    unordered_multiset<int> sample1, sample2, sample3;
 
    // List initialization
    sample1 = { 1, 2, 2, 3, 3, 4, 4, 4, 3, 4 };
    sample2 = { 1, 2, 3, 1, 4 };
 
    // Merge both unordered_multisets and
    // move the result to sample1
    sample3 = merge(sample1, sample2);
 
    // copy assignment
    sample1 = sample3;
 
    // Print the unordered_set list
    for (auto it = sample1.begin(); it != sample1.end(); ++it)
        cout << *it << " ";
    cout << endl;
 
    for (auto it = sample2.begin(); it != sample2.end(); ++it)
        cout << *it << " ";
    cout << endl;
 
    for (auto it = sample3.begin(); it != sample3.end(); ++it)
        cout << *it << " ";
    cout << endl;
}

Output:

1 1 1 2 2 2 3 3 3 3 4 4 4 4 4 
4 3 2 1 1 
1 1 1 2 2 2 3 3 3 3 4 4 4 4 4

Time Complexity-O(N)


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