Re: [理工] [OS]-multi level paging
※ 引述《bernachom (Terry)》之銘言：
: 請教一下..
: consider the following hardware configuration virtual address =32 bits,
: page size=4kbytes, and a page table entry occupies 4 bytes.
: How many pages should the OS allocate for the pages table of a 12Mbyte process
: under the following paging mechanisms?
: a)one-level paging
: b)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)
: ----
: a)這題我是這樣算，2^20 * 4 = 4MB, 12MB/4MB=3個page table
: b)算出來答案很怪...
: 請教一下，b)這題要怎麼算呢?
: 謝謝幫忙
※更正一下，換個想法看看
a) 12MB/4KB = 3K (代表這個process需要3K個page→代表PT需要3K個entry)
→再來計算 3K * 4byte = 12KB (代表這3K個entry佔12KB)
→12KB / 4KB = 3 (總共需要3個page)
b) 12MB /4KB =3K
→因為現在變成 2-level，所以第二層的PT只能貢獻2^10個entry
→由a知該Process需要3K個page
→3K / 2^10 = 3 (代表第二層需要3個page)
→還需要一個第一層對到第二層
→3 + 1 = 4
盡量寫的比較完整
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