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| '''Direct Mapped Cache''' | == Direct Mapped Cache == |
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https://www.scotnpatti.com/images/directmappedcache.jpg '''Tags''' contain the address information required to identify if the information in a cache location coresponds to the data needed. These tags contain the upper bits of the memory address. '''NOTE:''' The ''byte offset'' is the number of bits required to specify bytes in a block. Since this example uses 4 byte (word) sized blocks the ''byte offset'' is 2 bits https://www.scotnpatti.com/images/directmappedcache2.jpg '''To determine the size of the cache above: {{{ cache size = (number of locations=1024 = 2^10) X (Block size = 32 + Validity bit = 1 + (32 - TagSize=10 - Block bits = 2) = 2^10 X (32 + 1 + 32 - 10 - 2) = 1024 X (53) = 54,272 bits or 6,784 bytes }}} How many bits are required for a direct-mapped cache with 64 KB of data and one-word blocks, assuming a 32-bit address. {{{ 64/4 = 16 KWords = 2^16 Cache Size = 2^16 X (32 + (32 - 16 - 2) + 1) = 2^16 X 49 = 802816 bits = 100,352 bytes = 98 KB. }}} |
Cache
Direct Mapped Cache
Cache's are directed mapped if each memory location is mapped to exactly one location in the cache. An example would be:
location = (Block Address) MOD (Number of cache blocks in the cache)
In this way we can map a memory location to a cache location.
Example: Suppose we have a cache with 8 slots 2^3. Then a word at 45 would be found at slot 5 in the cache.
https://www.scotnpatti.com/images/directmappedcache.jpg
Tags contain the address information required to identify if the information in a cache location coresponds to the data needed. These tags contain the upper bits of the memory address. NOTE: The byte offset is the number of bits required to specify bytes in a block. Since this example uses 4 byte (word) sized blocks the byte offset is 2 bits
https://www.scotnpatti.com/images/directmappedcache2.jpg
To determine the size of the cache above: How many bits are required for a direct-mapped cache with 64 KB of data and one-word blocks, assuming a 32-bit address. cache size = (number of locations=1024 = 2^10) X (Block size = 32 + Validity bit = 1 + (32 - TagSize=10 - Block bits = 2)
= 2^10 X (32 + 1 + 32 - 10 - 2)
= 1024 X (53)
= 54,272 bits or 6,784 bytes64/4 = 16 KWords = 2^16
Cache Size = 2^16 X (32 + (32 - 16 - 2) + 1) = 2^16 X 49 = 802816 bits = 100,352 bytes = 98 KB.