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 --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.127.208.96