C++ map access discards qualifiers (const)

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 [].