(Same machine as above: 32-bit virtual address space, 4 KB page size.) Find the number of pages required for the given virtual address space.
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Correct answer: C. 2^20
Explanation
The number of pages in the virtual address space equals the total virtual address space size divided by the page size: 232 bytes (32-bit address space) ÷ 212 bytes (4 KB page) = 220 pages, matching option (iii) — this is the same count used as the number of page-table entries in the earlier page-table-size question.More Programming Languages and Computer Graphics questions
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