Iterator Question

[Yadrif]

Hello Everyone,
I would like to return a pointer to an object that is found in a set. Here is the code:

#include <cstdlib>
#include <iostream>
#include <string>
#include <set>

using namespace std;

class KeyValue
{
public:
    KeyValue(string key, string value) : _key(key), _value(value) {

    }

    string getName() const { return this-> _key; }

    string getValue() const { return this-> _value; }

    bool operator==(const KeyValue& other) const
    {
        return (this->_key == other.getName());
    }

    bool operator<(const KeyValue& other) const
    {
        return (this->_key < other.getName());
    }

private:
    string _key;
    string _value;
};

class KeyValueSet
{
public:

    KeyValueSet() {
        // put some KeyValue's in the set just for testing
        KeyValue kv1("Dog", "Hound");
        KeyValue kv2("Cat", "NoGood");
        KeyValue kv3("Fish", "Bass");
        theSet.insert(kv1);
        theSet.insert(kv2);
        theSet.insert(kv3);
    }

    KeyValue* getKeyValue(string & name, string & value)
    {
        KeyValue key_value(name, value);
        KeyValue * ret = 0;

        // find by name only, if not found use value as default
        set<KeyValue>::iterator iter = theSet.find(key_value);
        if (iter != theSet.end()) {
            //ret = iter; // compile error
            ret = const_cast<KeyValue*>(&(*iter)); // Is there a better way to do this?
        } else {
            ret = new KeyValue(name,value);
        }
        return ret;
    }

private:
    set<KeyValue> theSet;

};


int main(int argc, char *argv[])
{

    KeyValueSet kvs;

    string s1("Fish");
    string s2("Bass");
    KeyValue* kv = kvs.getKeyValue(s1, s2);

    system("PAUSE");
    return 0;
}

Is there an easier way to do this other than the line:
retval = const_cast<KeyValue*>(&(*iter));

I tried the line above it but get a compile error. I think I understand why after do some reading but it seems like there would be an easier way to return the pointer to the object.

All help is appreciated.

Thanks.