For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache.
Tag |
Index |
Offset |
31-10 |
9-4 |
3-0 |
How many entries does the cache have?
The index has 6 bits(9-4+1), and therefore the cache has 26= 64 entries.
Block size = 2Nwhere N = (number of offset bits-2) = 4-2 = 2. (Note that the “-2” is due to using 32-bit words and byte-addressed memory.) Block size = 22= 4 words
Alternately, block size = 2(number of offset bits)bytes.
Block size = 24= 16 bytes = 4 words
For a direct-mapped cache design with a 32-bit address, the following bits of the address are...
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