How did this loophole around const member function worked?

February 2, 2022

In the below code we try to multiply each element’s data in the list by 2 and assign it. But the apply function is a const function therefore should not be able to change the values of member fields. Output for the fifth line in main is

So code below succeeds in changing the values as intended and I can’t figure out why.

#include <iostream>
#include <list>
#include <string>

using std::ostream;
using std::cout;
using std::endl;

template<class E> class MyList {
class Node {
    friend class MyList<E>;
    E data;
    Node* next = nullptr;
  }; // end of class Node
  Node* head = new Node;
  Node* tail = head;
  MyList(const MyList&) = default;
public:
  MyList() = default;
  MyList& operator=(const MyList&) = delete;
  void push_front(const E& data) {
    Node* node = new Node;
    node->data = data;
    node->next = head->next;
    head->next = node;
    if(head->next == nullptr) tail = node;
  }
  void push_back(const E& data) {
    if(head->next == nullptr) {
        push_front(data); return;
    }
    MyList temp(*this);
    temp.head = temp.head->next;
    temp.push_back(data);
    temp.head = nullptr;
  }
  ~MyList() {
    Node *node = head, *next;
    while(node != nullptr) {
        next = node->next;
        delete node;
        node = next;
    }
  }
  template<class Function>
  void apply (Function f) const {
    Node* node = head->next;
    while(node != nullptr) {
        f(node->data);
        node = node->next;
    }
  }
};


int main() {
    MyList<int> m1;
    m1.push_back(3);
    for(int i = 1; i <= 2; ++i) m1.push_front(i); 
    for(int i = 1; i <= 2; ++i) m1.push_back(i); 
    m1.apply(
        [](auto& val){ val *= 2;}
    );
    m1.apply(
        [](const auto& val){cout << val << endl;}
    );
    return 0;
}

>Solution :

The key is logical vs bitwise constness. The head data member is a non-const pointer to non-const Node: the const correctness of the apply member function is bitwise constness: