Is it specified in the C++11 standard that std::begin(Container&&) returns const_iterator?

Question

Here's a link to relevant code:

#include <iostream>
#include <string>
#include <vector>
#include <type_traits>

int main()
{
  std::vector<int> v{1, 2, 3, 4, 5};
  auto iter = begin(std::move(v));
  if(std::is_const<typename std::remove_reference<decltype(*iter)>::type>::value)
    std::cout<<"is const\n";
  return 0;
}

http://coliru.stacked-crooked.com/a/253c6373befe8e50

I ran into this behavior because of a declval<Container>() in a decltype expression with std::begin . Both gcc and clang return iterators which yield const references when dereferenced. It probably makes sense since r-value references usually bind to expiring objects that you don't want to mutate. However, I could not find any documentation on this to determine whether it's mandated by the standard. I couldn't find any relevant overloads of begin() or ref-qualified overloads of Container::begin() .

Update: The answers clarified what's happening but the interactions can be subtle as demonstrated below:

#include <iostream>
#include <string>
#include <vector>
#include <type_traits>

int main()
{
  if(std::is_const<typename std::remove_reference<decltype(*begin(std::declval<std::vector<std::string>>()))>::type>::value)
    std::cout<<"(a) is const\n";
  if(!std::is_const<typename std::remove_reference<decltype(*std::declval<std::vector<std::string>>().begin())>::type>::value)
    std::cout<<"(b) is not const\n";
  if(!std::is_const<typename std::remove_reference<decltype(*begin(std::declval<std::vector<std::string>&>()))>::type>::value)
    std::cout<<"(c) is not const\n";
  return 0;
}

http://coliru.stacked-crooked.com/a/15c17b288f8d69bd

Naively, you wouldn't expect different results for (a) and (b) when ::begin is just defined in terms of calling vector::begin. However the absence of std::begin overloads that take non-const r-value reference and return iterator (or ref-qualified vector::begin overloads which return const_iterator) cause exactly that to happen.

Answer 1

As you can see in http://en.cppreference.com/w/cpp/iterator/begin the interesting overloads are:

template<class C> auto begin(C& c) -> decltype(c.begin());
template<class C> auto begin(const C& c) -> decltype(c.begin());

and std::vector<int>&& can only bind to the second overload (so returns const_iterator ).

Answer 2

Let's try to analyze what happens, step by step:

1. You're calling std::begin(std::vector<int>&&) , but std::begin has no overload that takes an rvalue :

    template< class C > auto begin( C& c ) -> decltype(c.begin()); template< class C > auto begin( const C& c ) -> decltype(c.begin()); 

---

2. Due to reference collapsing , a temporary (xvalue) will only bind to a const lvalue reference:

   If you call Fwd with an xvalue, we again get Type&& as the type of v. This will not allow you to call a function that takes a non-const lvalue, as an xvalue cannot bind to a non-const lvalue reference. It can bind to a const lvalue reference, so if Call used a const&, we could call Fwd with an xvalue.

   (From the linked answer) .

---

3. Therefore, the

     template<class C> auto begin(const C& c) -> decltype(c.begin()); 

   overload is being called, which returns a const iterator.

   Why?

   Because std::begin(v) calls v.begin() , which returns a const_iterator when called on const instances of std::vector .