Questions about a Direct-Mapped Cache

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Question II Consider a direct-mapped cache with 256 blocks and the block size is 16 bytes (or 4 words). Compute the tag, index, and byte offset of the address: 01FFF8AC16. (Note that the figure is not the cache described. It is only for showing a cache structure.)   Check solution

The 32-bit address is divided into three fields as follows:


• 4-bit byte offset field, because block size = 2⁴ = 16 bytes,


• 8-bit cache index, because there are 2⁸ = 256 blocks in cache, and


• 20-bit tag field, because 32 – 4 – 8 = 20.


Therefore, the three fields for the address 01FFF8AC₁₆ are


• Byte offset = C₁₆ = 12 (least significant 4 bits of the address),

Cache index = 8A16 = 138 (next lower 8 bits of the address), and Tag = 01FFF16 (upper 20 bits of the address).

Question III — Consider a cache with 32 blocks and a block size of 32 bytes. To what block number does the byte address 1600 map? Answer:   Block number: