Is capacity copied in a vector?

Question

Take the following code:

std::vector a;
a.reserve(65536);
std::vector b(a);  //NOTE: b is constructed from a

a.reserve(65536); // no r         


        

Answer 1

Standard says nothing about preserving capacity when you call copy constructor. So you have no any guarantees about it.

But you can do the following trick, which swap a's and b's state, if you need preserving capacity in the copy only:

 std::vector<int> a;
 a.reserve(65536);
 std::vector<int> b(a);
 b.swap(a); // now b has a's state
 assert(b.capacity() == 65536); 

Answer 2

1. As SGI STL vector soruce code shows below, operator= will reserve space for exactly n elements, i.e. _M_end_of_storage = _M_start + __xlen;.

    template <class _Tp, class _Alloc>
    vector<_Tp,_Alloc>&
    vector<_Tp,_Alloc>::operator=(const vector<_Tp, _Alloc>& __x)
    {
      if (&__x != this) {
        const size_type __xlen = __x.size();
        if (__xlen > capacity()) {
          iterator __tmp = _M_allocate_and_copy(__xlen, __x.begin(), __x.end());
          destroy(_M_start, _M_finish);
          _M_deallocate(_M_start, _M_end_of_storage - _M_start);
          _M_start = __tmp;
          _M_end_of_storage = _M_start + __xlen;
        }
        else if (size() >= __xlen) {
          iterator __i = copy(__x.begin(), __x.end(), begin());
          destroy(__i, _M_finish);
        }
        else {
          copy(__x.begin(), __x.begin() + size(), _M_start);
          uninitialized_copy(__x.begin() + size(), __x.end(), _M_finish);
        }
        _M_finish = _M_start + __xlen;
      }
      return *this;
    }

2. As SGI STL vector soruce code shows below, vector's copy constructor will reserve space for exactly n elements, i.e. _M_end_of_storage = _M_start + __n;.

      template <class _InputIterator>
      vector(_InputIterator __first, _InputIterator __last,
             const allocator_type& __a = allocator_type()) : _Base(__a) {
        typedef typename _Is_integer<_InputIterator>::_Integral _Integral;
        _M_initialize_aux(__first, __last, _Integral());
      }

      template <class _Integer>
      void _M_initialize_aux(_Integer __n, _Integer __value, __true_type) {
        _M_start = _M_allocate(__n);
        _M_end_of_storage = _M_start + __n;
        _M_finish = uninitialized_fill_n(_M_start, __n, __value);
      }

Answer 3

Is capacity copied?

In practice, no. I tested it online in Clang and GCC as well as MSVC and none of them copy the capacity.

Will there be a reallocation on the last line?

If the capacity is less than the argument to reserve (i.e. it doesn't get copied) then yes.

Does the standard say anything about this or is it silent?

No definitions for the copy constructor are provided in vector.cons. Instead we have to look at the container.requirements

X denotes a container class containing objects of type T, a and b denote values of type X, u denotes an identifier, r denotes a non-const value of type X, and rv denotes a non-const rvalue of type X.

X u(a)

X u = a;

Requires: T is CopyInsertable into X (see below).

post: u == a

Now what does it mean for two containers to be equal?

a == b

== is an equivalence relation. equal(a.begin(), a.end(), b.begin(), b.end())

In other words, since it doesn't require capacity to be equal in the comparison, then there's no reason to copy the capacity.

Answer 4

Well, a simple check like the following reveals that the capacity is not copied:

std::vector<int> a;
a.reserve(65536);
cout << "a.capacity is " << a.capacity() << endl; // prints 65536

std::vector<int> b(a);  //NOTE: b is constructed from a
cout << "b.capacity is " << b.capacity() << endl; // prints 0

I believe that upon copying vector a to b, the capacity of b is set to the size of a in most compilers; though this is not guaranteed.

Answer 5

No, capacity is not guaranteed to be preserved by a vector copy construction.

You can do that as follows:

vector<int> b;
b.reserve( a.capacity() );
b = a;

Better encapsulated in a function.