Homework Answers
Cache line or block size = 64 byte -> Block Offset = log 64 = 6 bits
Total # of block inside cache = 512/64 = 8
Therefore total # of set = 8/2= 4
Therefore index bits = log 4 = 2 bits
# of tag bits = 32 -(6+2) = 32-8=24 bits
|
Tag (24 bits) |
Set Index (2 bits) |
Block Offset (6 bits) |
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