※ 引述《work76 (work)》之铭言： : 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 12 20 page size=4kbytes = 2 bytes 12mbyte = 12*2 bytes 20 12 8 除以page size 12*2 / 2 = 12 * 2 = 3076 = 3k pages 每个page占一个entry 故 3k * 4bytes = 12 kbytes b. two level paging.:(assuming that the number of entries in a first leve page : table is the same as that in a second-level page table) : 可以帮忙算一下这题吗?也是完全不会的一题考古题，谢谢。 20 22 24 12mbyte = 12*2 bytes = 3 * 2 < 2 bytes(电脑要以2进位储存 找最接近 22 3*2 的2的倍数) 故取24 bits 扣到page size 的 12bit 剩12bit 由於要two level (题目说要拆两半)故12bit 拆成两半 第一个6bit 做first level 第2个6bit 做 second level table 6 first level=2 * 1(每个entry占1 byte 电脑最小分配资源) =64 bytes (first level指向second level的entry) 6 second level =2 * 4 bytes (指向记忆体位址 题目给的) = 256 bytes 故共 256 + 64 = 320 bytes (这是我的见解 可能有错) --

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