by Ciro Santilli (@cirosantilli, 34)
Page directory given to process by the OS:
entry index   entry address      page table address  present
-----------   ----------------   ------------------  --------
0             CR3 + 0      * 4   0x10000             1
1             CR3 + 1      * 4                       0
2             CR3 + 2      * 4   0x80000             1
3             CR3 + 3      * 4                       0
...
2^10-1        CR3 + 2^10-1 * 4                       0
Page tables given to process by the OS at PT1 = 0x10000000 (0x10000 * 4K):
entry index   entry address      page address  present
-----------   ----------------   ------------  -------
0             PT1 + 0      * 4   0x00001       1
1             PT1 + 1      * 4                 0
2             PT1 + 2      * 4   0x0000D       1
...                                  ...
2^10-1        PT1 + 2^10-1 * 4   0x00005       1
Page tables given to process by the OS at PT2 = 0x80000000 (0x80000 * 4K):
entry index   entry address     page address  present
-----------   ---------------   ------------  ------------
0             PT2 + 0     * 4   0x0000A       1
1             PT2 + 1     * 4   0x0000C       1
2             PT2 + 2     * 4                 0
...
2^10-1        PT2 + 0x3FF * 4   0x00003       1
where PT1 and PT2: initial position of page table 1 and page table 2 for process 1 on RAM.
With that setup, the following translations would happen:
linear    10 10 12 split  physical
--------  --------------  ----------
00000001  000 000 001     00001001
00001001  000 001 001     page fault
003FF001  000 3FF 001     00005001
00400000  001 000 000     page fault
00800001  002 000 001     0000A001
00801004  002 001 004     0000C004
00802004  002 002 004     page fault
00B00001  003 000 000     page fault
Let's translate the linear address 0x00801004 step by step:
  • In binary the linear address is:
    0    0    8    0    1    0    0    4
0000 0000 1000 0000 0001 0000 0000 0100
  • Grouping as 10 | 10 | 12 gives:
    0000000010 0000000001 000000000100
0x2        0x1        0x4
    which gives:
    page directory entry = 0x2
page table     entry = 0x1
offset               = 0x4
    So the hardware looks for entry 2 of the page directory.
  • The page directory table says that the page table is located at 0x80000 * 4K = 0x80000000. This is the first RAM access of the process.
    Since the page table entry is 0x1, the hardware looks at entry 1 of the page table at 0x80000000, which tells it that the physical page is located at address 0x0000C * 4K = 0x0000C000. This is the second RAM access of the process.
  • Finally, the paging hardware adds the offset, and the final address is 0x0000C004.
Page faults occur if either a page directory entry or a page table entry is not present.
The Intel manual gives a picture of this translation process in the image "Linear-Address Translation to a 4-KByte Page using 32-Bit Paging": Figure 1. "x86 page translation process"
Figure 1.
x86 page translation process
.

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