CODEWITHSUNDEEP
c++stlc++11unordered-map
Bee San
Bee San

Reputation: 2665

std::unordered_map and duplicate keys

I'm using an stl unordered_map, and I can't seem to get the count method to work. This is my program:

typedef unordered_map<char, int> Mymap;
int main() 
{
    Mymap m;  

    m.insert(Mymap::value_type('a', 1)); 
    m.insert(Mymap::value_type('b', 2)); 
    m.insert(Mymap::value_type('c', 3)); 
    m.insert(Mymap::value_type('b', 4)); 
    m.insert(Mymap::value_type('b', 5)); 

    cout << m.count('b') << endl;

    return 0; 
}

The documentation for unordered_map says that unordered_map::count(const Key& k) returns the number of elements with the key k. So I would expect the output here to be 3, whereas the real output is 1. Why?

Upvotes: 14

Views: 40697

Answers (2)

jfs
jfs

Reputation: 414149

// g++ -std=c++0x init-unorderedmap.cc && ./a.out
#include <iostream>
#include <unordered_map>

namespace {
  typedef std::unordered_map<char, int> Mymap;
}

int main() {
  using namespace std;

  Mymap m{ {'a', 1}, {'b', 2}, {'c', 3}, {'b', 4}, {'b', 5}};
  cout << m.count('b') << endl;

  unordered_multimap<char, int> mm{ {'b', 4}, {'b', 5}};
  cout << mm.count('b') << endl;
}

Output

1
2

Upvotes: 14

James McNellis
James McNellis

Reputation: 355039

An unordered_map maintains a 1:1 mapping of key to value, so count will always return zero or one.

You need an unordered_multimap if you want to map multiple values to a single key.

Upvotes: 38

Related Questions