Asked by abc
If you have a 16 MB logical address space and wish to have 8KB-pages for logical to physical mapping, how many bits should be allocated for indexing into the page table?
Answers
Answered by
oobleck
16M = 2^24
8K = 2^13
so you need 24-13 = 11 bits
8K = 2^13
so you need 24-13 = 11 bits
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