Does std::vector *have* to move objects when growing capacity? Or, can allocators "reallocate"?

Question

A different question inspired the following thought:

Does std::vector<T> have to move all the elements when it increases its capacity?

As far as I understand, the standard behaviour is for the underlying allocator to request an entire chunk of the new size, then move all the old elements over, then destroy the old elements and then deallocate the old memory.

This behaviour appears to be the only possible correct solution given the standard allocator interface. But I was wondering, would it make sense to amend the allocator to offer a reallocate(std::size_t) function which would return a pair<pointer, bool> and could map to the underlying realloc()? The advantage of this would be that in the event that the OS can actually just extend the allocated memory, then no moving would have to happen at all. The boolean would indicate whether the memory has moved.

(std::realloc() is maybe not the best choice, because we don't need do copy data if we cannot extend. So in fact we'd rather want something like extend_or_malloc_new(). Edit: Perhaps a is_pod-trait-based specialization would allow us to use the actual realloc, including its bitwise copy. Just not in general.)

It seems like a missed opportunity. Worst case, you could always implement reallocate(size_t n) as return make_pair(allocate(n), true);, so there wouldn't be any penalty.

Is there any problem that makes this feature inappropriate or undesirable for C++?

Perhaps the only container that could take advantage of this is std::vector, but then again that's a fairly useful container.

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Update: A little example to clarify. Current resize():

pointer p = alloc.allocate(new_size);  for (size_t i = 0; i != old_size; ++i) {   alloc.construct(p + i, T(std::move(buf[i])))   alloc.destroy(buf[i]); } for (size_t i = old_size; i < new_size; ++i) {   alloc.construct(p + i, T()); }  alloc.deallocate(buf); buf = p; 

New implementation:

pair<pointer, bool> pp = alloc.reallocate(buf, new_size);  if (pp.second) { /* as before */ } else           { /* only construct new elements */ } 

Answer 1

When std::vector<T> runs out of capacity it has to allocate a new block. You have correctly covered the reasons.

IMO it would make sense to augment the allocator interface. Two of us tried to for C++11 and we were unable to gain support for it: [1] [2]

I became convinced that in order to make this work, an additional C-level API would be needed. I failed in gaining support for that as well: [3]

Answer 2

In most cases, realloc will not extend the memory, but rather allocate a separate block and move the contents. That was considered when defining C++ in the first place, and it was decided that the current interface is simpler and not less efficient in the common case.

In real life, there are actually few cases where reallocis able to grow. In any implementation where malloc has different pool sizes, chances are that the new size (remember that vector sizes must grow geometrically) will fall in a different pool. Even in the case of large chunks that are not allocated from any memory pool, it will only be able to grow if the virtual addresses of the larger size are free.

Note that while realloc can sometimes grow the memory without moving, but by the time realloc completes it might have already moved (bitwise move) the memory, and that binary move will cause undefined behavior for all non-POD types. I don't know of any allocator implementation (POSIX, *NIX, Windows) where you can ask the system whether it will be able to grow, but that would fail if it requires moving.