std::deque memory usage - Visual C++, and comparison to others

Question

Follow up to What the heque is going on with the memory overhead of std::deque?

Visual C++ manages deque blocks according to the container element type

Answer 1

STLPort

... seems to use:

::: 
...
enum { _MAX_BYTES = 32 * sizeof(void*) };
...
::: 
...
static size_t _S_buffer_size()
{
  const size_t blocksize = _MAX_BYTES;
  return (sizeof(_Tp) < blocksize ? (blocksize / sizeof(_Tp)) : 1);
}

So that would mean 32 x 4 = 128 bytes block size on 32bit and 32 x 8 = 256 bytes block size on 64 bit.

My thought: From a size overhead POV, I guess it would make sense for any implementation to operate with variable length blocks, but I think this would be extremely hard to get right with the constant time random access requirement of deque.

As for the question

Does STL allow for overriding of this block size at compile-time, without modifying the code?

Not possible here either.

Apache

(seems to be the Rogue Wave STL version) apparently uses:

static size_type _C_bufsize () {
    // deque only uses __rw_new_capacity to retrieve the minimum
    // allocation amount; this may be specialized to provide a
    // customized minimum amount
    typedef deque<_TypeT, _Allocator> _RWDeque;
    return _RWSTD_NEW_CAPACITY (_RWDeque, (const _RWDeque*)0, 0);
}

so there seems to be some mechanism to override the block size via specialization and the definition of ... looks like this:

// returns a suggested new capacity for a container needing more space
template 
inline _RWSTD_CONTAINER_SIZE_TYPE
__rw_new_capacity (_RWSTD_CONTAINER_SIZE_TYPE __size, const _Container*)
{
    typedef _RWSTD_CONTAINER_SIZE_TYPE _RWSizeT;

    const _RWSizeT __ratio = _RWSizeT (  (_RWSTD_NEW_CAPACITY_RATIO << 10)
                                       / _RWSTD_RATIO_DIVIDER);

    const _RWSizeT __cap =   (__size >> 10) * __ratio
                           + (((__size & 0x3ff) * __ratio) >> 10);

    return (__size += _RWSTD_MINIMUM_NEW_CAPACITY) > __cap ? __size : __cap;
}

So I'd say it's, aehm, complicated.

(If anyone feels like figuring this out further, feel free to edit my answer directly or just leave a comment.)