Custom iterator in range based for: issue with constness - c++

Question

The following code is based on that found in Modern C++ programming cookbook, and is compiled in VS 2017:
#include <iostream>
using namespace std;
template <typename T, size_t const Size>
class dummy_array
{
T data[Size] = {};
public:
T const & GetAt(size_t const index) const
{
if (index < Size) return data[index];
throw std::out_of_range("index out of range");
}
// I have added this
T & GetAt(size_t const index)
{
if (index < Size) return data[index];
throw std::out_of_range("index out of range");
}
void SetAt(size_t const index, T const & value)
{
if (index < Size) data[index] = value;
else throw std::out_of_range("index out of range");
}
size_t GetSize() const { return Size; }
};
template <typename T, typename C, size_t const Size>
class dummy_array_iterator_type
{
public:
dummy_array_iterator_type(C& collection,
size_t const index) :
index(index), collection(collection)
{ }
bool operator!= (dummy_array_iterator_type const & other) const
{
return index != other.index;
}
T const & operator* () const
{
return collection.GetAt(index);
}
// I have added this
T & operator* ()
{
return collection.GetAt(index);
}
dummy_array_iterator_type const & operator++ ()
{
++index;
return *this;
}
private:
size_t index;
C& collection;
};
template <typename T, size_t const Size>
using dummy_array_iterator = dummy_array_iterator_type<T, dummy_array<T, Size>, Size>;
// I have added the const in 'const dummy_array_iterator_type'
template <typename T, size_t const Size>
using dummy_array_const_iterator = const dummy_array_iterator_type<T, dummy_array<T, Size> const, Size>;
template <typename T, size_t const Size>
inline dummy_array_iterator<T, Size> begin(dummy_array<T, Size>& collection)
{
return dummy_array_iterator<T, Size>(collection, 0);
}
template <typename T, size_t const Size>
inline dummy_array_iterator<T, Size> end(dummy_array<T, Size>& collection)
{
return dummy_array_iterator<T, Size>(collection, collection.GetSize());
}
template <typename T, size_t const Size>
inline dummy_array_const_iterator<T, Size> begin(dummy_array<T, Size> const & collection)
{
return dummy_array_const_iterator<T, Size>(collection, 0);
}
template <typename T, size_t const Size>
inline dummy_array_const_iterator<T, Size> end(dummy_array<T, Size> const & collection)
{
return dummy_array_const_iterator<T, Size>(collection, collection.GetSize());
}
int main(int nArgc, char** argv)
{
dummy_array<int, 10> arr;
for (auto&& e : arr)
{
std::cout << e << std::endl;
e = 100; // PROBLEM
}
const dummy_array<int, 10> arr2;
for (auto&& e : arr2) // ERROR HERE
{
std::cout << e << std::endl;
}
}
Now, the error is pointing at the line
T & operator* ()
stating
'return': cannot convert from 'const T' to 'T &'"
...which is raised from my range based for loop on arr2.
Why is the compiler choosing the none-constant version of operator*()?. I have looked at this for a long time; I think its because it thinks that the object on which it is calling this operator is not constant: this should be a dummy_array_const_iterator. But, this object has been declared to be constant via
template <typename T, size_t const Size>
using dummy_array_const_iterator = const dummy_array_iterator_type<T, dummy_array<T, Size> const, Size>;
...so I really don't understand what is happening. Can someone please clarify?
TIA

Answer 1

dummy_array_const_iterator::operator * should always return T const & regardless of constness of the iterator object itself.
The easiest way to achieve this is probably just to declare it with T const as underlying iterator value type:
template <typename T, size_t const Size>
using dummy_array_const_iterator = dummy_array_iterator_type<T const, dummy_array<T, Size> const, Size>;
Since you are returning the iterator by value its constness can be easily lost by c++ type deduction rules and just declaring dummy_array_const_iterator as alias to const dummy_array_iterator_type is not enough. i.e. the following fails:
#include <type_traits>
struct I { };
using C = I const;
C begin();
int bar()
{
auto x = begin(); // type of x is deduced as I
static_assert(std::is_same<I, decltype(x)>::value, "same"); // PASS
static_assert(std::is_same<decltype(begin()), decltype(x)>::value, "same"); // ERROR
}

Answer 2

I found a way to enable T& operator*() only when C is not constant:
template <class Tp = T>
typename std::enable_if<std::is_const<C>::value, Tp>::type const& operator* () const
{
return collection.GetAt(index);
}
template <class Tp = T>
typename std::enable_if<!std::is_const<C>::value, Tp>::type & operator* () const
{
return collection.GetAt(index);
}
I have no idea about the syntax (which I get from https://stackoverflow.com/a/26678178)