Question

Question 29 7 pts Suppose we have a byte-addressable computer with a cache that holds 8 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache block would the hexadecimal address Ox1F map if the computer uses direct mapping?

Answer 1

• Given byte addressable system
• main memory address = 8 bit // 8 bits needed to represent main memory address
• No of cache Blocks = 8 blocks = 2³blocks // 3 bits needed to represent no of cache blocks in cache memory
• Block size = 4 byte = 2²byte                                // 2 bits needed to represent block offset

• format of a memory address as seen by the Direct Map cache  
• tag =3 bits cache index = 3 bits offset = 2 bits

<..................................memory address 8 bit..............................................................>

Tag(3 bits)

Cache Block index (3 bits)

Offset ( 2 bits)

Direct Map cache

hexadecimal memory address 0x1F will map to cache block number 7 // binary (111 )₂

                                0x 1 F

0001 1111 // Now separate bits according to direct map format

000 1 11 11

• tag =000
• cache index = 111
• offset = 11

cache block No = (1 11)₂ = ( 7 )₁₀