`iterator` and `const_iterator` are not required members of STL containers?

Question

Taken from Understanding iterator/const_iterator implementation:

"although iterator and const_iterator are types declared in the scope of vector, there is no requirement that vector(or any STL container) have a member of either type - iterator and const_iterator are part of the interface of std::vector e.g. overloads of the member begin() returns those types, but nothing is said about how those function obtain the iterator they return"

Additionally STL containers must have:

"a begin and end function that returns iterators"

The above states that iterator and const_iterator are not required members of a STL container for example vector. I assume this means that the type returned from .begin or .end will differ based on implementation.

So I am wondering why this isn't problematic as I see a lot of people write out std::vector<someType>::iterator or std::vector<someType>::const_iterator where iterator and const_iterator are specified instead of using auto for example:

for (std::vector<int>::iterator i = s.begin(); i != s.end(); i++)
{

}

Answer 1

You're reading the quote wrong. The person says

have a member of either type

not

have either type

When they say have a member of either type they mean there is no data member of type iterator or const_iterator in the class.

They do go on to say

iterator and const_iterator are part of the interface of std::vector

Which is correct as the standard requires that std::vector surface these types in it interface.

Answer 2

std::vector will normally contain a typedef (or equivalently, using) to specify the type that the name iterator will refer to. At its simplest, it might be something like:

template <class T, class Allocator>
class vector {
    using iterator = T *;
    using const_iterator = T const *;
    // ...
};

Likewise, its begin and end must return iterators, and its cbegin and cend must return const_iterators. But, the vector object doesn't (necessarily need to contain any objects of type iterator or const_iterator.

Also note that although the names vector::iterator and vector::const_iterator need to be defined, it isn't strictly necessary that they be defined by vector itself. You could (for example) have vector derive from a base class defines the appropriate names:

template <class T>
class vector_base {
public:
    using iterator = T*;
    using const_iterator = T const *;
    // ...
};

template <class T, class Allocator>
class vector : public vector_base<T> {
};

So, even though vector itself doesn't define iterator, the name vector::iterator is well defined. I probably wouldn't bother to mention this except for one point: although it doesn't apply to vector, there is a standard class named std::iterator, that was intended primarily as a base class for iterators, and most of what it did was pretty much what's outlined above--define the member typedefs for value_type, difference_type, reference, and so on. Its use is now deprecated, but there are still a fair number of older iterator classes that used it (and older versions of the standard at least implied that most standard iterator types should use it as well).