Size of empty vector

Question

The following program on running with g++ 4.8.2 gave the output 12 on a 32-bit Linux system:

vector<char> v;
cout << sizeof(v) << endl;

I saw this and know that sizeof(v) could be implementation specific. Still, I was wondering what might be causing that vector to have a size of 12. What I think is that the iterators v.begin() and v.end() might be contributing to 8 bytes of the size. Am I correct? If yes, what is contributing to the remaining 4 bytes of size? If not, what are these 12 bytes all about?

Answer 1

Take a look at the sources. libstdc++ is part of the gcc download.

Anyway, the container must have these members:

1. A data-pointer, 4 bytes for a char*.
2. An element count or an end pointer, 4 bytes for a size_t or char*.
3. A buffer-size or pointer to end-of-buffer, 4 bytes for a size_t or char*.
4. The standard allocator (empty trivial type) needs no space, thanks to some implementation-tricks (Empty-baseclass-optimization, and perhaps partial template-specialization. C++20 could use the attribute [[no_unique_address]] instead).

In theory, 2 and 3 might be smaller if not pointers. Though that would be curious, as it would restrict the maximum size.

Also in theory, 2 and 3 could be allocated dynamically with the data. Haven't found anyone actually do that though.

Together 12 bytes, as expected.
Double the sizes for a 64-bit implementation.

Answer 2

Typically std::vector has:

1. Start of allocation / begin
2. End of vector (begin + size)
3. End of allocation (begin + capacity)

So size 12 is quite justified on a 32 bit machine.

Answer 3

libstdc++'s std::vector derived from a base with a data member of this type:

  struct _Vector_impl
  : public _Tp_alloc_type
  {
    pointer _M_start;
    pointer _M_finish;
    pointer _M_end_of_storage;
    ...

_M_end_of_storage supports .capacity() / resizing etc..