Is capacity copied in a vector?

Question

Take the following code:

std::vector<int> a; a.reserve(65536); std::vector<int> b(a);  //NOTE: b is constructed from a  a.reserve(65536); // no reallocation b.reserve(65536); 

Is capacity copied? Will there be a reallocation on the last line? Does the standard say anything about this or is it silent?

Answer 1

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 2

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);