对于具有 16 MB 可寻址位置的主存储器和每块 16 字节的 64 KB 直接映射缓存的系统，需要多少标记位。
(一) 4
(乙) 14
(三) 8
(四) 12答案： (C)
解释：
Main memory size = 16MB = 2^{24} byte
Cache memory size = 64KB = 2^{16} byte
Word size = 16 bytes = 2^{4} byte = 4 bits in WO.
Number of blocks
= Number of lines
= 2^{16} byte / 2^{4} byte = 2^{12} = 12 bits in LO.
Therefore, tag bits are = 24 - 12 - 4 = 8 bits in tag.
Option (C) is correct.Quiz of this Question
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