Get the first value (string) in a map from the second value (int)

Question

If you have a std::map which takes a string and an int.

    std::map<std::string, int> exampleMap;

Is there any way to print the string if you have the correct int?

Let's say I insert the string "hello" and the int 0 into my map:

    exampleMap.insert(std::make_pair("hello", 0));

To print 0, we can use:

    exampleMap.find('hello')->second;

Is there any way to print the "hello" string?

Answer 1

You can sequentially loop through all the entries in the map until you find the one with the value:

for (auto &it : exampleMap)
{
    if (it.second == value)
        return it.first;
}

Although you need to decide what to do if there are multiple entries with the same int value and also what to do if the int is not found.

Answer 2

Let's say I insert the string "hello" and the int 0 into my map:

 exampleMap.insert(std::make_pair("hello", 0));

Is there any way to print the "hello" string?

you need to iterate to find the entries having the expected value to print the key

#include <map>
#include <string>
#include <iostream>

int main()
{
  std::map<std::string, int> exampleMap;

  exampleMap.insert(std::make_pair("hello", 0));
  exampleMap.insert(std::make_pair("how", 1));
  exampleMap.insert(std::make_pair("are", 2));
  exampleMap.insert(std::make_pair("you", 0));

  int expected = 0;

  for (std::map<std::string, int>::const_iterator it = exampleMap.begin();
       it != exampleMap.end();
       ++it) {
    if (it->second == expected)
      std::cout << it->first << std::endl;
  }
}

Compilation and execution :

pi@raspberrypi:/tmp $ g++ -pedantic -Wall -Wextra m.cc
pi@raspberrypi:/tmp $ ./a.out
hello
you
pi@raspberrypi:/tmp $ 

Answer 3

You have a few choices...

1. Create two maps (one going from string to int and one going the other way)
2. You can use a "Bi-Directional" map, that will let you go both ways. Boost also has an implementation.
3. You can use a function like this one to convert from one map to another map.

Which one you choose depends on how often you need to do both lookups. If it is rarely, I'd use the conversion function. If it isn't rare, I would use the bi-directional container.