warning: converting to non-pointer type 'int' from NULL

Question

I need to write a dictionary in c++ There is a function that returns the value of a dictionary by key, but if there is no value, then I want to return something like NULL. The value 0 is not suitable because it could be the value of some other key. What can i use instead of 0?

TValue getByKey(TKey key)
{
    for (unsigned i = 0; i < this->pairs.size(); i++)
    {
        if (this->pairs[i].key == key)
            return this->pairs[i].value;
    }

    return NULL;
}

if I use NULL it gives me a warning (g++ compiler):

warning: converting to non-pointer type 'int' from NULL

Answer 1

NULL is specifically a value for pointers. If the object that you return is not a pointer (or a smart pointer), then returning NULL is wrong.

If you return an int, then you can only return a value that int can represent, i.e. 0 or 1 or 2 ...

If you want to return a value that represents "no value", then you can return instance of std::optional template instead. Example:

std::optional<TValue> getByKey(TKey key)
{
    // ...

    return std::nullopt;
}

---

P.S. You can replace your loop that implements linear search with the standard function std::find.

P.P.S If you have many values, then may want to consider using another data structure that has faster key lookup such as std::unordered_map.

P.P.P.S Don't use NULL for pointers either. It has been obsoleted by nullptr.