What will you see on the next line?
2 answers:
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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:
an error message
Explanation:
The return value is the value which is sent back by the function to a place in the code from where the was called from. Its main work is to return a value form the function.
In the context, the form of "(symbol-length? 'James)" in Scheme will return the value --- ' an error message'.
Answer:
There is also an attachment below
Explanation:
Since we are talking about binary search, let's assume that the items are sorted according to some criteria.
Time complexity of binary search is O(logN) in worst case, best case and average case as well. That means it can search for an item in Log N time where N is size of the input. Here problem talks about the item not getting found. So, this is a worst case scenario. Even in this case, binary search runs in O(logN) time.
N = 700000000.
So, number of comparisions can be log(N) = 29.3 = 29.
So, in the worst case it does comparisions 29 times
