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A machine has $48-bit$ virtual addresses and $32-bit$ physical addresses. Pages are $8\: KB.$ How many entries are needed for a single-level linear page table?
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$\because$ Total number of bits in the page number field = $48-12 \;\text{bits} = 36$

$\therefore 2^{24} = 16 \times 1024 \times 1024 \;\text{entries}$

by Boss (18.9k points)
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Why 12 bits?