I/O-mapping
Question 1 |
Consider a direct mapped cache with 64 blocks and a block size of 16 bytes. To what block number does the byte address 1206 map to
does not map | |
6 | |
11 | |
54 |
Question 1 Explanation:
Given data,
Direct memory cache have = 64 block
Block size = 16 Bytes
Block number does the byte address of 1206 map=?
Step-1: To find block number = Byte Address / Block size
= 1206/16
= 75.3
Step-2: Byte address 1206 map to 75th block.
Step-3: We have to find the cache block number.
Cache block number = (Block number) mod (Block size in cache)
= 75 mod 16
= 11
Direct memory cache have = 64 block
Block size = 16 Bytes
Block number does the byte address of 1206 map=?
Step-1: To find block number = Byte Address / Block size
= 1206/16
= 75.3
Step-2: Byte address 1206 map to 75th block.
Step-3: We have to find the cache block number.
Cache block number = (Block number) mod (Block size in cache)
= 75 mod 16
= 11
There is 1 question to complete.