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
The raw materials of photosynthesis, water and carbon dioxide, enter the cells of the leaf, and the products of photosynthesis, sugar and oxygen, leave the leaf.
Answer:
Basic
Explanation:
The 3rd generation programming language that most students learned when most computers used MS DOS was basic. It remains a safe programming language for using 3rd party code and provides a coding standard that integrates easily with other programming languages.
Answer:Computers only understand machine code - they do not understand high-level language code. Any high-level programming language code has to be converted to executable code. Executable code is also known as machine code which is a combination of binary code 0s and 1s.
Explanation:
void minMax(int a, int b, int c, int*big, int*small)
{
if(a>b && a >c){
*big = a;
if(b>c)
*small = c;
else
*small = b;
}
else if (b>a && b>c){
*big = b;
if(a>c)
*small = c;
else
*small = a;
}
else{
*big = c;
if(a>b)
*small = b;
else
*small = a;
}
}
