- 2^32 bytes of memory
- 2048 blocks (of 16 bytes each)
Now I’m trying to figure out how much bits each field will contain. Direct mapped: One block is 16 bytes (16 * 8 = 128 bits). The block also contains 1 dirty bit and 1 valid bit. I know that since there are 2048 (=2^11) blocks, and the whole block contains 16 bytes (=2^4) the tag-field will be 4 + 11 + 1 (valid) + 1 (dirty) = 17 bits. So the whole size of 1 block will be 17 + 128 bits. Is this right? And if so, where do I need the given 2^32 bytes of memory, and what is the difference with the direct-mapped cache in comparison to counting the 4-way-set-associative cache fields? I hope someone can help me, thanks in advance!
Asked By : Steven Raaijmakers
Answered By : TEMLIB
- address range : A[31:0]
- block address : A[3:0]
- cache line index : A[14:4] : 2048 = 11bits
Each tag contains :
- addresses A[31:15] : 17bits
- valid bit
- dirty bit, if it is a write-back cache
So 18bits (write-thu) or 19bits (write-back) per tag. 128+18 per block (or 128+19), 2048*(128+18) bits for the whole cache. (excluding EDC, sub-blocking…) The tags contains the part of the address bits not indexed by the cache (not the contrary). Way associative caches need replicating both tags and data. Additional bits are also needed in that case for the replacement information (LRU, PLRU…) : Selecting which cache way should be replaced when a cache miss occurs.
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