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. Thank you, Mrunal On Mon, Apr 21, 2008 at 8:44 AM, Nikhil Talpallikar < [EMAIL PROTECTED]> wrote: > Argh! my mistake. did the math in a hurry. > > > On Mon, Apr 21, 2008 at 9:06 PM, Luciano Rocha <[EMAIL PROTECTED]> > wrote: > > > On Mon, Apr 21, 2008 at 08:42:55PM +0530, Nikhil Talpallikar wrote: > > > Hi, > > > > > > Then should not the PAGE_SHIFT be 13 for the page size to be 4k on a > > x86 > > > arch.? > > > > Hm? 1 << 12 == 2^12 == 4096. > > > > -- > > Luciano Rocha <[EMAIL PROTECTED]> > > Eurotux Informática, S.A. <http://www.eurotux.com/> > > > >
