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Akimi4 [234]
2 years ago
15

What are the advantages and disadvantages of using a very large memory cell size, say, W = 64 instead of the standard size W = 8

? If each integer occupies one 64-bit memory cell and is stored using sign/magnitude notation, what are the largest (in terms of absolute value) positive and negative integers that can be stored? What if two cells are used to store integers?
Computers and Technology
1 answer:
stepladder [879]2 years ago
7 0

Answer and Explanation:

  • The benefits of an large size memory cell is that you can store a lot bigger numbers (and all information is, eventually, numbers) in one cell, which by and large improves the speed and unwavering quality of information get to.  
  • The inconvenience is that you go through a bigger bit of whatever information stockpiling medium you are utilizing on each expression of information, and accordingly can store less information generally speaking.  
  • The biggest positive integer that could be put away in a framework utilizing sign/size documentation, with 64-piece cells is 263. On the off chance that two such cells were utilized to store numbers, the biggest whole number that could be put away by the framework is 2127.
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Answer:

Hi There was small mistake. It is working fine for me. When you run from command line - use LoopPattern, not looppatern

import java.util.Scanner;

public class Looppattern {

  public static void main(String[] args) {

      Scanner sc = new Scanner(System.in);

      System.out.println("Enter how man levels you need: ");

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          for (int k = 1; k <= p; k++) { // increasing each level printing

              System.out.print(k);

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      int r = levels;

      for (int p = 1; p <= levels; p++) {

          for (int k = 1; k <= r; k++) {

              System.out.print(k);

          }

          r--; // decreasing levels

          System.out.println();

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      System.out.println("\n---------------Pattern C-----------------\n");

      for (int p = 1; p <= levels; p++) { // here incresing

          for (int k = p; k > 0; k--) { // and here decreasing pattern to

                                          // achieve our required pattern

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      System.out.println("\n---------------Pattern D-----------------\n");

      r = levels;

      for (int p = 1; p <= levels; p++) {

          for (int k = 1; k <= r; k++) {

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          }

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      }

  }

}

Explanation:

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Answer:

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Explanation: pls mark my answer as brainlist

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