Getting const_iterator from begin() instead of cbegin()

Question

Can someone explain why the following will not compile if I uncomment the line foo::const_iterator j = f.begin(); , but if I use the line foo::const_iterator j = f.cbegin(); it will compile? I'm trying to get that line to work just as it works with my std::vector example.

#include <vector>

struct foo {
    struct node { };
    node *first = nullptr, *last = nullptr;

    struct base_iterator {
        node* ptr;
        base_iterator (node* n) : ptr(n) { }
    };

    struct iterator : base_iterator { using base_iterator::base_iterator; };

    struct const_iterator : base_iterator { using base_iterator::base_iterator; };

    iterator begin() { return iterator(first); }
    const_iterator begin() const { return const_iterator(first); }
    const_iterator cbegin() const { return const_iterator(first); }
};

// Test

int main() {
    foo f;
    foo::iterator i = f.begin();
//  foo::const_iterator j = f.begin();  // Won't compile because f is not const.
//  foo::const_iterator j = f.cbegin();  // Will compile fine.

    std::vector<int> v;
    std::vector<int>::const_iterator it = v.begin();  // Compiles even though v is not const.
}

Answer 1

It works for std::vector because the iterators of all standard library containers are designed to support the iterator --> const_iterator conversion. It's meant to mimic how pointer conversions work.

Whenever your two iterators are user defined classes, you need to add it explicitly. You have two options:

A converting constructor:

struct iterator : base_iterator { using base_iterator::base_iterator; };

struct const_iterator : base_iterator {
   using base_iterator::base_iterator;
   const_iterator(const iterator& other) : base_iterator(other) {}
};

A conversion operator:

struct const_iterator : base_iterator { using base_iterator::base_iterator; };

struct iterator : base_iterator {
  using base_iterator::base_iterator; 
  operator const_iterator() const { /* ... */ }
};