Victor wrote:
struct Item
{
int *pPtr;
std::list<int*> intList;
};
Item *pItem = new Item[1700000];
I expected the memory usage would be (sizeof(Item) * 1700000) +
(the head pointer of list * 1700000), ie. (28+12)*1700000 = ~64MB,
but the outcome is about 170MB. Would you tell me why is it so?
No, not really. Do you run in debug mode? Using _SECURE_STL to
trace the iterators?
The std::list implementation might also allocate a head node, which
also requires memory.
Also, an allocation of 12 bytes might actually become 16 bytes on
some systems - to reduce heap fragmentation.
I asked this question in the MSDN forums and someone answered me
"You should not only account for the array itself (28 x 1700000 =
47.6M) but also for the size of each constructed std::list object.
That's two nodes, 2 x 36 bytes in your case. That adds up to
47.6M + 2 x 36 x 1700000 = 170M."
Don't know how they came up with that.
But according to my understanding
list()
: _Mybase(), _Myhead(_Buynode()), _Mysize(0)
{ // construct empty list
}
The default contructor will only allocate memory for _Myhead using
"_Nodeptr _Pnode = this->_Alnod.allocate(1);" which allocates only
12 bytes (_Nodeptr size). Could you let me know which two nodes
will be allocated in the contructor of std::list?
Also, is there any method to reduce the memory usage of a
std::list or I need to write my own link list?
One obvious way of reducing memory usage is to NOT have a
std::list of int pointers. The overhead is pretty big, compared to
the pointers themselves. Would a std::vector or a std::deque do
the job better?
Bo Persson