Given 64 bit logical space, Page Size = $4KB$. Do the three level paging?
My Solution:-
1. Since address space is =$ 2^{64} Bytes$
Page size = $2^{12}$ Bytes.
Number of entries in third level Page table is :- $2^{64} / 2^{12}$ = $2^{52}$.
Size of third level Page table = $2^{52} * 4B = 2^{54}$.
Number of entries in second level page table is = $2^{54} / 2^{12} = 2^{42}$. Size of second level page table = $2^{44}$.
Likewise in last level = $2^{44} / 2^{12} = 2^{32}$.
Size of 1st level page table = $2^{34}$.
It's correct right ?
Doubt :- They write corresponding bits for indexing page tables at three levels as:-
| $1st$ level |
$2nd$ level |
$3rd$ level |
Offset |
| $ 32$ |
$10$ |
$10$ |
$12$ |
Please explain how did they find this out ?
Please :)
