Re: [理工] [計組]-清大93-page計算
※ 引述《polomoss (小澤)》之銘言:
: Consider a logical address of 2048 pages of 1024 words (4byte) each,
: mapped onto a physical of 64 frames. Assume the smallest memory
: allocation unit is one byte
: a. How many bits in the logical and physical address
a. physical:
64frame*1024word*4byte=2^18 byte
physical至少要有18bit才能夠標出全部的記憶體位置
(allocation unit is one byte)
logical:
由上題得知index部分為2^12
2048page=2^11
logical至少要有11+12=23bit來標明
: b. one-level page table is used, how many bytes are required by PT
b.
pagetableㄧ個entry裡面含有
1.vaild bit 2.dirty bit 3.reference bit 4.physical address(部分)
共有2048個page===>2048個entry
2048entry*(3(for vaild ,dirty,reference)+6(phycial扣掉index))
= 9*2^11bit = 9*2^8byte
: c. two-level PT is used, and first level table has 32 entries,
: what is the minimal amount of memory required by PT.
c.
32=2^5
32entry要標完2048個page==> 2048/32=2^6 L2至少要有2^6個entry
L1: 32entry*(3+6(for L2))=9*2^5bit=9*2^2byte
L2: 2^6entry*(3+6(for phycial address))=9*2^6bit=9*2^3byte
L1+L2=108byte
: 可以幫我算一下嗎?(以及列式)
: 另外題目有點看不懂,是說總共2048個page?
: 每個page有1024個word? 那(4byte)又是指什麼?
1個word有4byte
: 謝謝
不一定正確 有錯誤請指正
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.38.100.229
推
01/10 23:31, , 1F
01/10 23:31, 1F
討論串 (同標題文章)
本文引述了以下文章的的內容:
完整討論串 (本文為第 2 之 7 篇):