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  • C++: Why does gcc prefer non-const over const when accessing operator[]?

    - by JonasW
    This question might be more appropriately asked regarding C++ in general, but as I am using gcc on linux that's the context. Consider the following program: #include <iostream> #include <map> #include <string> using namespace std; template <typename TKey, typename TValue> class Dictionary{ public: map<TKey, TValue> internal; TValue & operator[](TKey const & key) { cout << "operator[] with key " << key << " called " << endl; return internal[key]; } TValue const & operator[](TKey const & key) const { cout << "operator[] const with key " << key << " called " << endl; return internal.at(key); } }; int main(int argc, char* argv[]) { Dictionary<string, string> dict; dict["1"] = "one"; cout << "first one: " << dict["1"] << endl; return 0; } When executing the program, the output is: operator[] with key 1 called operator[] with key 1 called first one: one What I would like is to have the compiler choose the operator[]const method instead in the second call. The reason is that without having used dict["1"] before, the call to operator[] causes the internal map to create the data that does not exist, even if the only thing I wanted was to do some debugging output, which of course is a fatal application error. The behaviour I am looking for would be something like the C# index operator which has a get and a set operation and where you could throw an exception if the getter tries to access something that doesn't exist: class MyDictionary<TKey, TVal> { private Dictionary<TKey, TVal> dict = new Dictionary<TKey, TVal>(); public TVal this[TKey idx] { get { if(!dict.ContainsKey(idx)) throw KeyNotFoundException("..."); return dict[idx]; } set { dict[idx] = value; } } } Thus, I wonder why the gcc prefers the non-const call over the const call when non-const access is not required.

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