A direct mapped cache holds 8 16-word blocks. What is the correct breakdown of t

Question

A direct mapped cache holds 8 16-word blocks. What is the correct breakdown of the bits in a 32 bit address? 2 bits for the byte displacement 3 bits for the block displacement 4 bits for the cache index 23 bits for the tag 2 bits for the byte displacement 4 bits for the block displacement 2 bits for the cache index 24 bits for the tag 2 bits for the byte displacement 4 bits for the block displacement 3 bits for the cache index 23 bits for the tag 2 bits for the byte displacement 2 bits for the block displacement 0 bits for the cache index 28 bits for the tag

Explanation / Answer

c.) 2 bits for the byte displacement   

   4 bits for the block displacement

   3 bits for cache index

   23 bits for the tag

The reason behind this is as follows:

size of cache=32 KB =25* 23Bytes=28 Bytes.

No of Tag bits = 32 - 8 =23.

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