Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line = 
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
Answer:
35
Explanation:
We will be going inside B array, because he was in second place in array E,and will be the first element of array B, so it's 35
E can be understanded as:
E=[[21, 'dog', 'red'],[35, 'cat', 'blue'],[12, 'fish', 'green']], so, you can see array E as array of arrays or so-called two-dimensional array
Answer:
provides the platform that application software runs on, manages a computer's hardware, and implements features like file and folder management.
Explanation: