→ Reylod: 512GB = 2^39B => 39 bit address, 2^52 / 2^12 = 2^40 12/01 16:15
→ yad50968: 不好意思 想問一下 4kb pages 是指數量 那2^52/2^12 12/01 16:33
→ yad50968: 為什麼是代表最大的數量 12/01 16:33
→ Reylod: 每個page 4KB = 2^12B 12/01 16:43
Given a computer system with a 52-bit virtual address , 4kb pages, and 4
bytes per page entry.
Suppose that the maximum physical memory size is 512GB, and the system is
byte-addressable.
Let paging be implemented for the system.
(a)What is the number of bits for physical addresses ,and(b) what is the
maximum number of pages for a process.
ANS:
a. 39
b. 2^40
a = 39 是因為 實體記憶體位置有0~2^39-1 所以要39bit嗎?
還有請問b小題是怎麼解的 不知道為什麼會到2^40
謝謝!!
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