Question

Using direct mapping, consider a 16-bit memory addresses, and a cache with 64 blocks, where each block is 8 bytes. What is the size of offset, block, and tag

Answer 1

Tag Block Offset

7 6 3

Step-by-step explanation:

Using direct mapping hashing:

Memory address size = 16 bits

Number of cache blocks = 64

Size of each block = 8 bytes

Thus;

Offest field = size of each block = 2^{3} = 3 bits

Number of blocks = 64 = 2

Thus, the index field is = 2^{6} = 6 bits of block

Tag filed is of size = 16 - (3+6) = 7 bits of tag