Consider a machine with a byte addressable main memory of 2^16 (65536) bytes, wh

Question

Consider a machine with a byte addressable main memory of 2^16 (65536) bytes, which has a direct-mapped cache with 32 lines. Lines are 16 bytes long. How is the 16-bit memory address partitioned by the cache? (In other words, how big is the tag field, the entry into the line, etc.)? Into what line would bytes with each of the following addresses be stored? 0x111B 0xC334 0xD01D 0xAAAA Suppose the byte with address 0x2B2B is stored in the cache. What are the addresses of the other bytes stored along with it? How many total bytes of memory can be stored in the cache? Why is the tag also stored in the cache?

Explanation / Answer

a)  

  

Memory

Address

                

index

b) 0xC334

c)101011

d)16 bytes of memory

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

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