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This 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
      |                       |
      .                       .
      |                       |
      +-----------------------+  0
Notice 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

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  1. Example: simplified single-level paging scheme
  2. x86 Paging Tutorial
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