How to find out the size of std::list?

[stereoMatching]

my os : windows xp sp3()32bits
compilers : gcc4.5(minGW)

  std::list<size_t> haha;
  haha.resize(10);
  std::iota(haha.begin(), haha.end(), 0); //initialize haha to 0~9 

I want to find out how many memory allocated for each node
2 pointers plus one size_t, it should be 4 + 4 + 4 = 12bytes
But I expect it would be more than 12bytes
How could I find out the size of the list occupy by the default allocator?
Thanks a lot

[jsmith]

In general you can't know that unless you want to write your own allocator that counts. Even if you knew how much space the allocator allocates, you should understand that the C++ runtime may allocate more anyway. For example:

#include <iostream>

int main()
{
    int* p = new int;
    int* q = new int;
    std::cout << p << ' ' << q << std::endl;
}

outputs 0x9867008 0x9867018 for me. Subtracting the two pointer values yields 0x10, or 16 decimal, despite the fact that sizeof(int) == 4. This is because the minimum heap block size is 16 bytes. Every new will consume at least 16 bytes of memory, even if you only allocate 1 byte.

[stereoMatching]

Thanks a lot, maybe someday I have to implement my allocator
But looks like boost already do that for us?