[理工] 100 暨大資工 os memory

[jordanforme (jordan)]

virtual address is 32bits
2 level paging
first level paging uses 6 bits
second level paging uses 12 bits

(a)how many enties are there in the first level paging table
and in each of the secong level paging table?

這題應該就是first level 有 2^6個entry
            second level 有2^12個entry
應該對吧？

(b)assuming that each entry occupies 4 bytes,estimate the
total memory used for all the second level paging?

2^12 * 2^2 = 2^14 bytes

這樣算對嗎？
不知道有沒有我想的那麼簡單
題目寫各佔10分
不知道有沒有陷阱
謝謝！

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推 A4P8T6X9:我覺得第二題要在乘2^6，總共有2^6個level 2 每一個有2^1
→ A4P8T6X9:2個entry，又每個entry 4 byte，所以總共level 2 是2^6*2
→ A4P8T6X9:^12*2^2。
→ jordanforme:thanks!