作者gn00618777 (123)
看板Grad-ProbAsk
标题[理工] [OS]-政大资科98
时间Wed Feb 3 14:01:47 2010
政大资科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)
看了之前人家问的..我还是没有很清楚
(2) user program有3k个pages,而level-2 page table 可存1K个pages
所以 3k/1k = 3个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
问题一： 整个page table size有包含level-1 table吗？
问题二： 3*4是代表什麽..？
问题三： 整个问题是不是在问你建立12MB process的PT需要几张page?
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