Ciro Santilli
OurBigBook.comThis is how the memory could look like in a single level paging scheme:
Links Data Physical address
+-----------------------+ 2^32 - 1
| |
. .
| |
+-----------------------+ page0 + 4k
| data of page 0 |
+---->+-----------------------+ page0
| | |
| . .
| | |
| +-----------------------+ pageN + 4k
| | data of page N |
| +->+-----------------------+ pageN
| | | |
| | . .
| | | |
| | +-----------------------+ CR3 + 2^20 * 4
| +--| entry[2^20-1] = pageN |
| +-----------------------+ CR3 + 2^20 - 1 * 4
| | |
| . many entires .
| | |
| +-----------------------+ CR3 + 2 * 4
| +--| entry[1] = page1 |
| | +-----------------------+ CR3 + 1 * 4
+-----| entry[0] = page0 |
| +-----------------------+ <--- CR3
| | |
| . .
| | |
| +-----------------------+ page1 + 4k
| | data of page 1 |
+->+-----------------------+ page1
| |
. .
| |
+-----------------------+ 0Notice that:
- the CR3 register points to the first entry of the page table
- the page table is just a large array with 2^20 page table entries
- each entry is 4 bytes big, so the array takes up 4 MiB
- each page table contains the physical address a page
- each page is a 4 KiB aligned 4KiB chunk of memory that user processes may use
- we have 2^20 table entries. Since each page is 4KiB == 2^12, this covers the whole 4GiB (2^32) of 32-bit memory