Finding value in unordered_map

Question

I am using Boost unordered_map. I have a key value pair for each entry. How could I determine whether a particular value exist in the map? (I don\'t want to create another u

Answer 1

How about the following:

typedef std::unordered_map<int,std::string> map_type;
typedef std::unordered_map<int,std::string>::value_type map_value_type;

map_type m;

if (m.end() != find_if(m.begin(),m.end(),[](const map_value_type& vt)
                                           { return vt.second == "abc"; }
                                           ))
   std::cout << "Value found." << std::end;
else
   std::cout << "Value NOT found." << std::end;

Or using an external variable that is captured:

std::string value = "abc";
if (m.end() != find_if(m.begin(),m.end(),[&value](const map_value_type& vt)
                                                 { return vt.second == value; }))
   std::cout << "Value found." << std::end;
else
   std::cout << "Value NOT found." << std::end;

Answer 2

Why can't we use count method instead of find()

Description: Count elements with a specific key Searches the container for elements whose key is k and returns the number of elements found. Because unordered_map containers do not allow for duplicate keys, this means that the function actually returns 1 if an element with that key exists in the container, and zero otherwise.

unordered_map<int, int> hash;
    //converted array into hashMap
    for(int i=0; i<6; i++)
    {
        hash[i];
    }

    //commom elemenest value is set to 1 in hashMap
    for(int i =0; i<7; i++)
    {
        //element exist in array1
        if(hash.count(i))
        {
            hash[i] = 1;
        }
    }

Answer 3

Boost has the Bimap, which is a bidirectional map (ie, keys and values both refer to each other). This sounds more appropriate for your needs than the unordered_map.

Answer 4

You need to iterate over all the elements in the unordered_map and see if the given value exists.

The std::find_if algorithm with a custom predicate can be used to simplify this.