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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