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puteri [66]
3 years ago
6

For the preceding simple implementation, this execution order would be nonideal for the input matrix; however, applying a loop i

nterchange optimization would create a nonideal order for the output matrix. Because loop interchange is not sufficient to improve its performance, it must be blocked instead. (a) What should be the minimum size of the cache to take advantage of blocked execution

Computers and Technology
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

hi your question lacks the necessary matrices attached to the answer is the complete question

1024 bytes

Explanation:

A) The minimum size of the cache to take advantage of blocked execution

 The minimum size of the cache is approximately 1 kilo bytes

There are 128 elements( 64 * 2 ) in the preceding simple implementation and this because there are two matrices and every matrix contains 64 elements .

note:  8 bytes is been occupied by every element therefore the minimum size of the cache to take advantage of blocked execution

= number of elements * number of bytes

= 128 * 8 = 1024 bytes ≈ 1 kilobytes

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Adding transitions to slides in a presentation.
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Create a structure representing a student. The member variables should include student name, student ID, and four test grades fo
Neporo4naja [7]

Answer:

Output:

Name: Brainly

ID:0001

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0001

10 9 8 10

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Explanation:

#include<iostream>

#include<string>

using namespace std;

//variables declaration

struct Student {

  string id; //string declaration ID

  string name; // string declaration name

  int grades[4]; //array of 4 because it is 4 grades

};

//definition of the function get information

void inputData(Student &s){

 

   

   cout << "Name:" ;

   getline(cin,s.name);

   cout << "ID:";

   cin >> s.id;

   cout << "Write the 4 tests grades of the student separated by space :";

   for (int i = 0; i<4; i++)

       cin >> s.grades[i];

}

//definition of the function of average

double inputAvg(Student s){

   double summation;

   int temporary;

   double average;

   for (int i = 0; i<4; i++){  

      for (int j = i; j<4; j++){

         if (s.grades[j] > s.grades[i]){

             temporary = s.grades[i];

             s.grades[i] = s.grades[j];

             s.grades[j] = temporary;

         }

      }

    }

    summation = 0;

    for (int i = 0; i<3; i++){

        summation = summation + s.grades[i];  

    }

    average = summation/3;

    return average;

}

void disp(Student *s){

   cout << s->name << endl;

   cout << s->id << endl;

   for (int i = 0; i<4; i++)

       cout << s->grades[i] << " ";

   cout << endl;

   cout << "Average :" << inputAvg(*s) << endl;

   

}

int main(){

  Student st;

  int ch;

  while(true){

      inputData(st);

      disp(&st);

      cout << " '1' to continue '0' to exit :";

      cin >> ch;

      if (ch == 0)

         break;      

  }

  return 0;

}

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import math

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