Question

Consider this situation:

void doSmth1(std::map<int,int> const& m);

void doSmth2(std::map<int,int> const& m) {
  std::map<int,int> m2 = m;
  m2[42] = 47;
  doSmth1(m2);
}

The idea is that doSmth2 will call doSmth1 and forward the map it received from its caller. However, it has to add one additional key-value pair (or override it if it is already there). I would like to avoid copying the whole thing just to pass an additional value to doSmth1.

Answer 1

You can’t do that with the standard map. But if your problem is that specific, you might consider passing the new element separately:

void doSmth1(std::map<int, int> const & m, int newkey, int newvalue);

void doSmth2(std::map<int, int> const & m)
{
    doSmth1(m, 42, 47);
}

---

Update: If you really just want one map, and copying the map is out of the question, then here’s how you can implement @arrowdodger’s suggestion to make a temporary modification to the original map:

void doSmth2(std::map<int, int> & m)
{
    auto it = m.find(42);

    if (it == m.end())
    {
        m.insert(std::make_pair(42, 49));
        doSmth1(m);
        m.erase(42);
    }
    else
    {
        auto original = it->second;
        it->second = 49;
        doSmth1(m);
        it->second = original;
    }
}