Question

The following code says that passing the map as const into the operator[] method discards qualifiers:

#include <iostream> #include <map> #include <string>  using namespace std;  class MapWrapper { public:     const int &get_value(const int &key) const {         return _map[key];     }  private:     map<int, int> _map; };  int main() {     MapWrapper mw;     cout << mw.get_value(42) << endl;     return 0; } 

Is this because of the possible allocation that occurs on the map access? Can no functions with map accesses be declared const?

MapWrapper.cpp:10: error: passing const std::map<int, int, std::less<int>, std::allocator<std::pair<const int, int> > > as this argument of  _Tp& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](const _Key&)  [with _Key = int, _Tp = int, _Compare = std::less<int>,  _Alloc = std::allocator<std::pair<const int, int> >] discards qualifiers 

Answer 1

std::map‘s operator [] is not declared as const, and cannot be due to its behavior:

T& operator[] (const Key& key)

Returns a reference to the value that is mapped to a key equivalent to key, performing insertion if such key does not already exist.

As a result, your function cannot be declared const, and use the map’s operator[].

std::map‘s find() function allows you to look up a key without modifying the map.

find() returns an iterator, or const_iterator to an std::pair containing both the key (.first) and the value (.second).

In C++11, you could also use at() for std::map. If element doesn’t exist the function throws a std::out_of_range exception, in contrast to operator [].