This program searches for a user entered word in a dictionary text file and outputs what line it is on. I need to count the number of comparisons made to find the word in both a linear and binary search. Right now it says there were zero comparisons made. Any ideas on where or how to implement these counters would be much appreciated.
string linearSearch(const vector<string> &L, string k);
string binarySearch(const vector<string> &L, string k, int a = 0, int b = -1);
int count1 = 0;
int count2 = 0;
int main()
{
    ifstream inFile;
    inFile.open("dictionary.txt");
    vector < string > words;
    string line;
    while (getline(inFile, line))
    {
        words.push_back(line);
    }
    inFile.close();
    string userWord;
    cout << "Search for a word: " << endl;
    cin >> userWord;
    if (words.empty())
    {
        return -1;
    }
    cout << "Using binary search, the word " << userWord << " is in slot "
            << binarySearch(words, userWord) << ". There were " << count2
            << " comparisons  made." << endl;
    cout << "Using linear search, the word " << userWord << " is in slot "
            << linearSearch(words, userWord) << ". There were " << count1
            << " comparisons made." << endl;
    return 0;
}
string linearSearch(const vector<string> &L, string k)
{
    for (int i = 0; i < L.size(); ++i)
    {
        count1++;
        if (L[i] == k)
        {
            count1++;
            return to_string(i);
        }
    }
    return to_string(-1);
}
string binarySearch(const vector<string> &L, string k, int a, int b)
{
    ++count2;
    if (b == -1)
    {
        b = L.size();
    }
    int n = b - a;
    if (n == 0)
    {
        return to_string(-1); //?
    }
    int mid = (a + b) / 2;
    if (L[mid] == k)
    {
        ++count2;
        return to_string(mid);
    }
    else if (L[mid] > k)
    {
        ++count2;
        return binarySearch(L, k, a, mid);
    }
    else
    {
        count2 += 2;
        return binarySearch(L, k, mid + 1, b);
    }
    return to_string(-1);
}
 
     
    