Consider a logical-address of 2048 pages of 1024 words (4 bytes) each,
mapped onto a physical memory of 64 frames. Please answer the following
questions assuming that the smallest memory allocation unit is one byte
(8 bits).
(c)Assuming a 2-level page table is used and the first-level table has 32
entries, what is the minimal amount of memory (in bytes) required by the
page tables?
可以計算出每個page=1kb 共有2048個pages 以1-level來計算
table size=2048*6(entries)bits 以上one-level的算法沒錯吧?
(64 frames=2^6)
再來是2-level的所有table size p1=5 bit p2=6 bit
再算第一level的table size時 我就卡住了到底它的每個entry大小為何呢?
且如何計算第一level的table size大小呢呢?
謝謝
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