Consider a machine with a byte addressable main memory of 216 bytes and block si

Question

Consider a machine with a byte addressable main memory of 216 bytes and block size of 8 bytes. Assume that a direct mapped cache consisting of 32 lines is used with this machine. 1. How is a 16-bit memory address divided into tag. line number, and byte number? 2. Into what line would bytes with each of the following addresses be stored?? 3. Suppose the byte with address 0001 1010 0001 1010 is stored in the cache. What are the addresses of the other bytes stored along with it? 4. How many total bytes of memory can be stored in the cache?? 5. Why the tag is also stored in the cache??

Explanation / Answer

1.

8 leftmost bits = tag; 5 middle bits = line number; 3 rightmost bits = byte number.
TAG = 8

LINE = 5

WORD= 3

2.

Line 3.
Line 6.
Line 3.
Line 21.

3.

Bytes with addresses 0001 1010 0001 1000 through 0001 1010 0001 1111 are stored in the cache.

4.

256 bytes.

5.

Because two items with two different memory addresses can be stored in the same place in the
cache. The tag is used to distinguish between them.

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