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Lena [83]
3 years ago
6

A two-way set-associative cache has lines of 16 bytes and a total size of 8 kB. The 64-MB main memory is byte addressable. Show

the format of main memory addresses.
Computers and Technology
1 answer:
Dennis_Churaev [7]3 years ago
4 0

There are 64 bytes in one cache. Since there are 256 bits for data.Thus,there are 256 sets of 2 lines in a cache memory each. total 8 bits are required to identify the number of set number.For the 64 Mbyte main memory, a 26-bit address is needed. For the 64-MB main memory which consists of 2^22 cells.Therefore, the bits for set plus tag length would be 22 bits, tag length is 14 bits and the word field length is 4.

Explanation:

1 cache = 64 bytes

16 bits for "tag" and 256 bits (32 bytes) for data.

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Answer:

public class Solution {

   public static void main(String args[]) {

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       String searchString = "cHRis";

       

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       for(String name : listOfName){

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       for(String result : resultName){

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Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

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The second for-loop goes through the elements of the resultName array and output it. The output is empty if no element was found added to the resultName variable.

During comparison, the both string were converted to lower case before the comparison because same lowercase character does not equal same uppercase character. For instance 'A' is not same as 'a'.

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