Question

There is the class A containing two overloaded methods getItems();

typedef std::vector <int> TItems;

template <typename T>
class A 
{
private:
    T a;
    TItems items;

public:
    A(){}
    A ( const T a_, const TItems & items_) : a(a_) , items (items_) {}
    bool operator () ( const A <T> &aa ) {return a < aa.a;}
    TItems const & getItems() const {return items}
    TItems & getItems() {return items}
};

and the set of A objects

template <typename T>
struct TSet {typedef std::set <A <T> > Type;};

I would like to return const reference / reference to TItems, but only the second method works

int main ()
{
TSet <double> ::Type t;
TSet <double> ::Type::iterator it = t.begin();
t.insert (A <double>( 5, TItems(10,10)));

const TItems *items = &(it->getItems()); //OK
TItems *items = &(it->getItems()); //Error  
}

Error   1   error C2440: 'initializing' : cannot convert from 'const TItems *' to 'TItems *

Is it the reason that non-constant references enables to modify A objects causing a potential rearrangement od the set? But items of the set are not arranged by A.items but by a.

Is there a way how modify A.items using a non-constant reference?

Answer 1

Is it the reason that non-constant references enables to modify A objects causing a potential rearrangement od the set?

Exactly. std::set‘s elements (and BTW std::map‘s keys) are immutable, the structure will only give you const-qualified elements. So you have the option

• change your structure to std::map and put your a as key and items as data
• if you are absolutely sure you won’t break the ordering by manipulation with items, you can const_cast (or declare items mutable if that suits you).