Sistemas Operativos, Manejo de la memoria 9 Segmentación combinada con paginación — Transcript

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1. 0:00Hi, how are you? How are you doing?
2. 0:03Welcome to the Operating Systems
3. 0:05Fundamentals course. We are looking at
4. 0:08memory management. I am referring to
5. 0:10chapter 8 of the Silberschatz and
6. 0:12Galvin textbook. Now we are going to
7. 0:16look at segmentation with paging. We
8. 0:20have already seen what paging is, we
9. 0:22have also seen what segmentation is,
10. 0:24and now we will see that with this, we
11. 0:26can take advantage of both segmentation
12. 0:28and paging. We will start with the
13. 0:36Multics address translation scheme,
14. 0:37where the problems of external
15. 0:39fragmentation and long search times are
16. 0:41resolved by paging the segments. That
17. 0:46is, a segment will be paged, it will be
18. 0:48divided into pages, and each segment
19. 0:54contains the base address for each page
20. 0:56table. Okay? Let's see how it is here.
21. 1:01I have a logical address that is formed
22. 1:03by a segment and an offset. Remember
23. 1:08that we must add the segment table base
24. 1:10register to the segment to know where
25. 1:12the segment table begins. Now, what
26. 1:16will I have? Well, with this, I will
27. 1:21obtain the segment length and the page
28. 1:23table base, okay? And for this offset,
29. 1:29what I am going to do first is compare
30. 1:31it with the segment length to see if it
31. 1:33does not exceed the segment size. If it
32. 1:38exceeds the segment size, this should
33. 1:39cause an exception. If not, then this
34. 1:45offset is divided into two parts, which
35. 1:47are page and offset. Okay? So, what is
36. 1:52going to happen? Well, we are going to
37. 1:56add the base of the page table that is
38. 1:58here in the segment entry to this page.
39. 2:05In the segment table, we add this to
40. 2:07obtain the frame in the page table for
41. 2:08that segment. Okay? Okay. So, once we
42. 2:16obtain the frame, we add the offset to
43. 2:19this frame, and now I have a physical
44. 2:21address. which is formed by a frame and
45. 2:27an offset. Okay? Well, now let's see
46. 2:36how it is done with Intel systems, okay
47. 2:38? How is that address translation done?
48. 2:45What happens is that Intel uses
49. 2:46segmentation with paging for memory
50. 2:48management with two levels of paging.
51. 2:59In Intel systems, the segment address
52. 3:01is known or called a selector, and this
53. 3:03selector will help us search a segment
54. 3:05descriptor table to find the descriptor
55. 3:07for that segment, that is, where that
56. 3:09segment begins. All the segment data is
57. 3:16there, okay? in the segment descriptor.
58. 3:21Okay. I will add the offset to the
59. 3:27initial segment address, and this will
60. 3:29form a linear address composed of two
61. 3:31paging levels where I have the page
62. 3:33directory, right? And with this, I will
63. 3:41know in which page directory entry I
64. 3:43have to look for the page table. And in
65. 3:53the page, I will know how much I have
66. 3:54to move in this directory to know which
67. 3:56entry contains the frame where the
68. 3:58memory address I want to access is.
69. 4:08Finally, well, I have the offset, which
70. 4:10is within the frame, at which memory
71. 4:11address within the frame the address I
72. 4:13am looking for is. Right? Okay. Thank
73. 4:21you very much for your attention. I
74. 4:23hope it was clear, and if not, well, we
75. 4:25can ask questions. M.