操作系统导论-较小的表

本文最后更新于:2022年5月12日 晚上

第二十章:分页:较小的表

  1. 对于线性页表,只要知道第一个Page的地址,存于寄存器当中,就可以通过这个地址依次陆续推算下一个或后面任意一个有效的地址范围。对于多级页表,通过多次搜索,依旧可以在只有最初的页表的地址的情况下,通过多次的偏移查询来定位到最后需要的特定地址。

  2. 这里取例子说明算法,具体答案通过-c参数可直接输出

    这里以seed0的时候为例

    首先,在README.md中可以得到以下信息

    • Page Size: 32 bytes
    • Virtual Address Space:32 KB(1024个分页 2^15)
      • 虚拟地址需要 15 bits (VPN 占 10 bit,offset 占 5 bit)
    • Physical Memory: 128个分页(2^12)
      • 物理地址需要 12 bits (PFN 占 7bit, offset 占 5 bit )
    • Virtual Address Space 的前五位对应了Page Directory的索引

    通过seed生成0对应的地址数据,用于PDE查表,内容如下

    Page Content
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    page   0:1b1d05051d0b19001e00121c1909190c0f0b0a1218151700100a061c06050514
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    并且根据模拟生成的PDBR: 108可以知道Page 108对应的内容即是第一级PDE的映射

    接下来就可以进行计算了,这里取一个vaildinvaild的答案分析计算过程

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    Virtual Address 611c:
    --> pde index:0x18 [decimal 24] pde contents:0xa1 (valid 1, pfn 0x21 [decimal 33])
    --> pte index:0x8 [decimal 8] pte contents:0xb5 (valid 1, pfn 0x35 [decimal 53])
    --> Translates to Physical Address 0x6bc --> Value: 08
    Virtual Address 3da8:
    --> pde index:0xf [decimal 15] pde contents:0xd6 (valid 1, pfn 0x56 [decimal 86])
    --> pte index:0xd [decimal 13] pte contents:0x7f (valid 0, pfn 0x7f [decimal 127])
    --> Fault (page table entry not valid)

    Virtual Address 611c

    1. 将611c按二进制转换为Virtual Address Space对应的 15 bits 为 11000 01000 11100。对PDE分割前五位有11000(decimal 24)

    2. 这个时候对最顶层的PDTR对应的Page 108进行查表操作

    首先可以通过Page Content得到Page 108内容如下

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    Page 108: 83 fe e0 da 7f d4 7f eb be 9e d5 ad e4 ac 90 d6 92 d8 c1 f8 9f e1 ed e9 a1 e8 c7 c2 a9 d1 db ff
    Index: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

    通过查表找到对应Index的内容可以知道24(11000)对应位置的内容是a1,转换为二进制为1010 0001,读取最高位为1可以知道是vaild。再通过后面的0010 0001转换为十进制为33,因此再继续查询Page 33的内容

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    Page 33:7f 7f 7f 7f 7f 7f 7f 7f b5 7f 9d 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f 7f f6 b1 7f 7f 7f 7f
    Index: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

    611c的[1-5]位为PDE,这时我们则需要通过[6-10]位来通过PTE寻找VPN对应的PFN地址。由611c11000 01000 11100可以知道这次为01000(decmial 8),对照上面page 33可以发现第八个对应的内容为53,因此可以知道最终VPN对应上了PFNPage 53。最后再通过offset11100,配合Page 53对应的二进制PFN地址11 0101结合offset11100110101 11100,也就是最后答案的0x6bc,其中通过offsetPage 53中查表可以找到对应的 Value08

    Virtual Address 3da8

    1. 611c同理,将3da8转为二进制01111 01101 01000,得到pde index01111(15),对应的Value0xd6(1101 0110),通过1101 0110最高位可以知道目标分页为vaild,去掉vaild bit可以得到索引的pte0x56(0101 0110 | 86)

    2. 检索Page 86,由PTE index01101可以找到对应的Value0x7f(0111 1111)由于最高位为0,因此是invaild,也就是说pte无效,无法访问

    其他的情况也是依次类推,即可算出结果是否有效

  3. 根据个人理解,具体缓存hitmiss的概率主要由对应内存上的数据决定。对于访问次数较多的内容hit缓存加快速度的概率自然更高;而对于不重复的多次查询(比如上面的练习),自然miss的概率会更高。因此对于进程来说,如果能将数据尽量集中在某一块分页上则能有效提高内存访问和处理的速度。


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