Answer:
int main()
{
int x,y;
bool b1, b2;
cout<<"enter values of x : ";
cin>>x;
cout<<"enter values of y : ";
cin>>y;
b1 = y > x; // false
b2 = x > y; // true
cout << "b1 is = " << b1 << "\n";
cout << "b2 is = " << b2 << "\n";
if (b2)
cout<<endl << "Larger Number" << "\n";
else
cout << "Smaller Number" << "\n";
return 0;
}
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