Which bits of the address would be used in the tag, index and offset in a two-way set associative cache with 1-word blocks and a
total capacity of 512 words
1 answer:
Size of block = 1 word = 8 bytes
number of blocks = size of cache / size of word
= 512 word /1 word = 512 blocks
In a two-way set associative cache, 2 blocks will form 1 set. So number of index = 512/2=256=2^8
Size of offset = 3 bits
size of index = 8 bits
Size of tag = 64-11=53 bits
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I'm assuming hardware.
Remember, Hard = Physical.
If that's not it, try peripherals.
