[理工] OS page table

[ken1325 (優質水瓶男)]

Virtual address = 32 bits, page size=4k bytes
and a page table entry occupies 4 bytes.

How many pages should the OS allocate for the page tables of a 12M byte

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)


請問這兩題要怎麼算？

題目問有幾個pages，是指page table的entry有幾個嗎？
還是問有幾個page table？

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推 A4P8T6X9:page table所佔的page個數
推 int5566:(1) 12M/4K = 3K,3K*4/4K=3(pages)
→ int5566:3K是process縮需的page數,4是phy addr佔的的空間,4K是
→ int5566:1個page可以存的資料量
→ int5566:(2)2-level的是切成有1024張1024個phy addr的page table
→ int5566:所以3K個phy addr需要3張page table
→ A4P8T6X9:還要一個level 1唷。
→ int5566:(10+1024*10)*3/8/4K=1    有錯請指正謝謝
→ int5566:上面的2個10分別是第一層的1格跟第二層的1格的大小
推 int5566:oh no 我第二個算法有問題...我再回一篇好了