推 zeowo:標題錯誤 01/21 22:37
Assume a page size of 8KB and a page table entry takes 4 bytes,how many levels
of page tables would be required to map 64-bit address space if the top-level
page table fits into a single page?
page size = 8KB =2^13bytes
64-13=51bits
2^X*4byte=8KB
2^x+2 =2^13
所以x=11
level=51/11=5
ans:5
實在是不太為什麼 2^X*4byte=8KB 這行
希望大家幫我解答一下囉^^
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