Set associative cache

  • Thread starter hholzer
  • Start date
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If I have the addressing format:

Tag: 31-12
Index: 11-6
Byte offset: 5-0

And if my cache size is 32KB = 32*2^10 = 2^5*2^10 = 2^15

then this would represent a how many-way set associative
cache? i.e., an x-way associative cache?

Since bits 11-6 are used for the index, then this would
mean that there are 6 bits used for the number of sets:
2^6 sets but how many blocks per set? If I have
determined this correctly, it would be:
2^6 = (2^15)/(x*2^6) or 2^3 = x ?

So it would represent an 8-way set associative cache.
Then, the number of words in a block would be:
2^6/2^2 = 2^4? That doesn't seem right to me.
 

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