Use the source, Alex. In particular, have a look at include/linux/mm.h.
This is a back-of-the-envelope calculation:
For every page, a mem_map_t (struct page) is used to describe it.
I haven't followed this thread, so let's assume this is a 32 bits machine
with pages of 4Kb. You then need 48 bytes for every 4096 bytes of RAM.
For 2059720K, you will have a mem_map of 24137K or something.
If you exclude some memory regions, you get 23528k :-).
Cheers,
Marnix Coppens
--- Reality is that which | Artificial Intelligence when you stop believing | stands no chance against in it doesn't go away. (Philip K. Dick) | Natural Stupidity.
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