[理工] [計組]關於virtual memory
consider the following hardware configuration . virtual address=32bit, page
size=4kbytes, and a page table entry occupies 4 bytes. how many pages should
the os allocate for the pages tables 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)
page size=4KB
每個page entry的資料=4byte
a.
12M/4K = 3K(全部所需page entry數)
3K*4byte(page總容量)
3K*4/4K=3 #
b.
page offset=page bit數=12bit(因為4K=2^12)
page number=32-12=20
因L1=L2，所以L1 page num=10，L2 page num=10。
即page entry數=2^10=1K
12M/4K = 3K
3K/1K=3(L2 page table)
然後需用到1個L1的page table去查詢L2的table
所以總page數=3(L2)+1(L1)=4 #
上面是答案 = = 到底page是要算什麼阿? entry 和page 不一樣嗎?
還有b答案的最後不是算page table數目嗎? 不懂????
不知哪那大大可以解決疑惑 感恩!! orz..
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※ 編輯: happykeny 來自: 61.227.138.253 (02/11 19:17)
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