Answer:
When designing a cache, you have to consider this things:
If the cache has a bigger block size may have a lower delay, but when miss the miss rate will be costly. If an application has high spatial locality a bigger block size will do well, but programs with poor spatial locality will not because a miss rate will be high and seek time will be expensive.
Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
i believe the Nintendo 64
Answer:
-3874₁₀ = 1111 1111 1111 1111 1111 1111 1101 1110₂
Explanation:
2's complement is a way for us to represent negative numbers in binary.
To get 2's complement:
1. Invert all the bits
2. Add 1 to the inverted bits
Summary: 2's complement = -N = ~N + 1
1. Inverting the number
3874₁₀ = 1111 0010 0010₂
~3874₁₀ = 0000 1101 1101₂
2. Add 1 to your inverted bits
~3874₁₀ + 1 = 0000 1101 1101₂ + 1
= 0000 1101 1110₂
You can pad the most signigicant bits with 1's if you're planning on using more bits.
so,
12 bits 16 bits
0000 1101 1110₂ = 1111 0000 1101 1110₂
They asked for double word-length (a fancy term for 32-bits), so pad the left-most side with 1s' until you get a total of 32 bits.
32 bits
= 1111 1111 1111 1111 1111 1111 1101 1110
Are you trying to change the shape of an item within your program by entering the values into text boxes that become variables?
What language are you using?