On 29.05.2012 11:25, Era Scarecrow wrote:
On Tuesday, 29 May 2012 at 06:56:40 UTC, Dmitry Olshansky wrote:
On 29.05.2012 10:52, Era Scarecrow wrote:
I considered that, but then you actually limit your address space to
2^63,
No you don't. Since pointer is already a pointer to word-sized object.
It has 2 last bits == 0. Always. There is no escaping of this fact.
And no your address space is intact. All it has to do is assuming
proper alignment, and you sure have it since you _allocate_ it.
To be more specific most allocator go even farther and provide 8bytes
aligned pointer.
Don't you mean the first 2 bits? (being least significant). Even so,
that seems more like its working out of coincidence, that or you can
never use more than 25% of your memory in a single address space in your
program (ever).
Yes, least significant. no you memory is intact.
I think it's a classical case of you not getting how it works.
No problem, bear with me :)
Suppose you have array of size_t, let size_t.sizeof == 4.
Here it is:
data: a0 a1 a2 a3 a4 ...
addr: 20 24 28 32 36 ...
Noticed the pattern? Once first address is multiple of 4 (that is
properly aligned) it can't become not a multiple of 4.
Meaning that 2 least significant bits are always == 0.
The memory is not wasted since address is incremented by 4 bytes at a
time, and is read by a unit of 4 bytes.
--
Dmitry Olshansky
