On Tue, Apr 22, 2008 at 12:25 AM, Mrunal Gawade <mrunal.gawade@xxxxxxxxx> wrote:
Size of (page) = 4kb ( 2^12 ) (on X86 arch)
32 bit address space available == 2^32
therefore number of pages (virtually) available == 2^32/2^12 == 2^20
so we have 2^20 pages or page numbers or page addresses available.
now consider we have a 32 bit address XXXX.....
right shift it by 12 bits, voila! i have the page number.
got it :)
cheers,
Nikhil
This is in continuation of above query.
Virtual address is = 32 bit - 10 + 10 + 12 bits
10 bits -> Page directory entry offfset
10 bits -> Page table entry offset
12 bits- > Offset of data within page
---------------
I read somewhere that PAGE_SHIFT macro gives the page number of the page. Now if we consider this 32 bit address, and shift of 12 bits to the left how is that justified. I feel confused.
Size of (page) = 4kb ( 2^12 ) (on X86 arch)
32 bit address space available == 2^32
therefore number of pages (virtually) available == 2^32/2^12 == 2^20
so we have 2^20 pages or page numbers or page addresses available.
now consider we have a 32 bit address XXXX.....
right shift it by 12 bits, voila! i have the page number.
got it :)
cheers,
Nikhil
