Computers and Technology, 21.06.2020 01:57 wypozworski

Consider a byte-addressable main memory consisting of 4 blocks, and a cache with 2 blocks, where each block is 4 bytes.
This means Block 0 and 2 of main memory map to Block 0 of cache, and
Blocks 1 and 3 of main memory map to Block 1 of cache.
Using the tag, block, and offset fields, we can see how main memory maps to
cache as follows.
First, we need to determine the address format for mapping. Each block is 4
bytes, so the offset field must contain 2 bits; there are 2 blocks in cache, so the
block field must contain 1 bit; this leaves 1 bit for the tag (as a main memory
address has 4 bits because there are a total of 24
=16 bytes).

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Consider a byte-addressable main memory consisting of 4 blocks, and a cache with 2 blocks, where ea...

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