CODEWITHSUNDEEP
c++c++11stl
Vahagn
Vahagn

Reputation: 139

The std::pair in the std::map is returned as const

My example code is

class A
{
    int a = 0;
public:
    void setA(const int value)
    {
        a = value;
    }
};

std::map<std::string, std::set<A>> Map{{"A", {}}};
Map.rbegin()->second.rbegin()->setA(2);

I get the following error: "Member function 'setA' not viable: 'this' argument has type 'const A', but function is not marked const"

My question is why rbegin() returns a const pointer to A? Or why is the std:pair's second a const in the std::map?

Upvotes: 0

Views: 952

Answers (2)

Edgar Rokjān
Edgar Rokjān

Reputation: 17483

Basically, rbegin() returns a reverse iterator which points to an object of type A which is stored in the std::set in a const manner.

The reason behind such behaviour is quite simple: it is necessary to protect std::set from inadvertent changes of elements which are stored inside.

You should remember, that std::set stores its elements in a tree-like data structure to ensure fast search/insert/remove operations. Possible changes of elements inside std::set might lead to wrong elements comparison and corruption of data structure, that is why all iterators return by begin()/end() methods and their analogues expose elements in a const fashion.

Upvotes: 2

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385204

All std::set elements are exposed in const fashion. That's because they're both keys and values, and if you could modify keys willy-nilly then you'd ruin the tree structure inside the set.

It is currently not possible to directly modify set elements. You will have to remove then re-insert.

(This has nothing to do with the encapsulating map.)

Upvotes: 3

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