Why does std::array::begin return a pointer instead of an iterator?

Question

I am trying to build a nested iterator template and relied on iterators having various traits like value_type. But as it turns out, not all STL types even return iterators with those traits. For instance:

#include <array>
#include <type_traits>

template <typename T>
using iterator_t = decltype(std::declval<T>().begin());

static_assert(std::is_same_v<iterator_t<std::array<int, 3>>, int*>);

This code compiles and shows that the actual type of the array iterator is int*. In that case, how can I still access traits such as value_type etc?

You can access traits using std::iterator_traits<iterator_t>. They are specialized for pointer types. — Piotr Skotnicki
– Piotr Skotnicki, Commented May 26, 2020 at 10:33

Jan Schultke · Accepted Answer · 2024-01-09 08:44:12Z

8

The standard doesn't specify how the iterator should be implemented and what the exact type it should be. It just says that std::array is a ContiguousContainer, meaning that its iterator is ContiguousIterator.

In fact, pointers like int* do satisfy the requirements of the iterator of std::array, so it's quite legitimate for implementation.

You can use std::iterator_traits to get the value_type, like:

std::iterator_traits<std::iterator_t<std::array<int, 3>>>::value_type

It works with pointers too.

edited Jan 9, 2024 at 8:44

Jan Schultke

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answered May 26, 2020 at 10:35

songyuanyao

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Bathsheba · Accepted Answer · 2020-05-26 10:34:43Z

6

int* is an iterator as it satisfies all the requirements necessary of an iterator.

Informally:

You can deference it (unless it's past the final element of the array).
You can increment it (as it's a pointer in an array).
You can copy int* to another int*.
You can call std::swap with int* as the types.

answered May 26, 2020 at 10:34

Bathsheba

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