Question

In: Electrical Engineering

Suppose we have a direct-mapped cache that can hold a total of 1024 blocks with 4...


Suppose we have a direct-mapped cache that can hold a total of 1024 blocks with 4 words per block.

Compute the block index, block offset, and the tag for the following addresses:
(a) 0x11001001
(b) 0x00010014
(c) 0x01000004
(d) 0x01001018
(e) 0x7bdcca10

Solutions

Expert Solution

Answer :- Since 1024 blocks, hence number of index bits = 10. Two bits for offset since 4 words per block.

Thus fro LSB, 2-bit gives the value of block offset, next 10 bits gives the value of block index and rest bits are the value of tag.

a) 0x11001001 => 0001_0001_0000_0000_0001_0000_0000_0001,
block offset = 01, block index = 0000_0000_00, and tag = 0001_0001_0000_0000_0001.

b) 0x00010014 => 0000_0000_0000_0001_0000_0000_0001_0100,
block offset = 00, block index = 0000_0001_01, and tag = 0000_0000_0000_0001_0000.

c) 0x01000004 => 0000_0001_0000_0000_0000_0000_0000_0100,
block offset = 00, block index = 0000_0000_01, tag = 0000_0001_0000_0000_0000.

d) 0x01001018 => 0000_0001_0000_0000_0001_0000_0001_1000,
block offset = 00, block index =  0000_0001_10, tag = 0000_0001_0000_0000_0001.

e) 0x7bdcca10 => 0111_1011_1101_1100_1100_1010_0001_0000,
block offset = 00, block index = 1010_0001_00, tag = 0111_1011_1101_1100_1100.

Manojponduru answered 2 years ago
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