You have a K-Way set associative cache with following parameters Cache Data Size (C) 2048
Block Size (b) 16
Blocks/Ways per set (K) 4

After partitioning the address, which is 32 bits big, into Tag, Set, and Offset, how many bits will be in each field?

Following are the solution to the given question:

Explanation:

The size of cache  memory

The block size

Therefore, are blocks.

The collection consists of 4 blocks. Therefore,

are sets. Because the size of its cache block is 16, the block offset comprises 4 bits. There would be 5 bits inset offset since there are 32 caches. An Address contains 32 bits, while the space of the tags contains bits.

i'm pretty sure data is stored in a form