Question: Suppose a computer using direct-mapped cache has 2 20 bytes of byte-addressable main memory and a cache of 32 blocks, where each cache block contains
Suppose a computer using direct-mapped cache has 2 20 bytes of byte-addressable main memory and a cache of 32 blocks, where each cache block contains 16 bytes
a) How many blocks of main memory are there?
b) What is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, block, and offset fields?
c) To which cache block will the memory address 0x0DB63 map?
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a To find the number of blocks in main memory we can divide the total size of main memory by the blo... View full answer
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