7.A memory system has 1M words. Each word is an 8-bit byte. The memory is
organized into blocks of 8 words each. The cache has 256K words, organized
into cache lines of 8 words each. The memory cache is organized into 4-way
set associative cache.
A.13 bits are needed for cache set address.
B.4 bit are for Tag.
C.21 bits are needed for address all words.
D.17 bits are needed for address all blocks.
E.None of the above
【解答】A, B
Physical address bit:23 bits(223=8*1M), Block offset bit:6 bits(26=8*8).
Cache line Number=256K/8=32K, Set Number=32K/4=8K, Index bit:13 bits.
Tag bit:23-6-13=4 bits.
20 bits are needed for address all words, 15 bits(215=32K) are needed for
address all blocks.
這解答是 徐老寫的 跟 張凡的書有不同的地方在於 張凡是Physical address bit:20
我在想 他們可能對於 Each word is an 8-bit byte 有不同解釋 是吧?
張凡老師的答案 多給個D .
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.193.80.151
※ 編輯: BitTorrent 來自: 123.193.80.151 (01/28 20:08)