Question

A physical address is 32 bits and the virtual address is split as 10 bits ‘off-set’,2 bits ‘byte addressability’, 20 bits being left for identifying the page.

What size are the pages and the page tables?

I believe its:

2^10 = 1024.

The ones that are reserved pages are 2^2 = 4

I am not sure on how to use the addressbility though…

Thanks again guys 🙂

Answer 1

let me tell you:

In the architecture that you use, you will use 2²⁰ pages, and offset will help you identify the word location in a page. So page size is 2¹⁰. What is the 2 bits of addressability? It is the addressability of bytes in a word => you have 4 bytes in a word and these 2 bits maps to the byte locations in your word.