政大資科98
Consider the following hardware configuration. Virtual address=32bit
page size = 4kbyte, and a page table entry occupies 4 bytes. How many
pages should the OS allocate for the pages tables of a12Mbyte process
under the following paging mechanisms?
1 one-level paging
2 two-level paging(Assuming that the number of entries in a first-level
page table is the same as that in a second-level
page table)
看了之前人家問的..我還是沒有很清楚
(2) user program有3k個pages,而level-2 page table 可存1K個pages
所以 3k/1k = 3個level-2 page table
page table size = 2^10*2^2*3 + 3*4=(12*2^10+12) bytes
需要(12*2^10+12)/2^12 =4 pages
問題一： 整個page table size有包含level-1 table嗎？
問題二： 3*4是代表什麼..？
問題三： 整個問題是不是在問你建立12MB process的PT需要幾張page?
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※ 編輯: gn00618777 來自: 220.138.105.41 (02/03 14:55)
※ 編輯: gn00618777 來自: 220.138.105.41 (02/03 21:18)