(Same machine as above: 16 GB main memory, 32-bit virtual address space, 4 KB page size.) What is the size of the page table needed to handle the given virtual address space, given that each page table entry is of size 2 bytes?
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Correct answer: C. 2 MB
Explanation
The virtual address space is 32 bits, so the number of pages (page-table entries needed) is 232 / 212 (page size 4 KB = 212 bytes) = 220 entries. Each page-table entry is 2 bytes, so the page table size = 220 × 2 bytes = 221 bytes = 2 MB, matching option (iii) (assuming a single flat page table covering the whole virtual address space).More Programming Languages and Computer Graphics questions
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