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grin007 [14]
3 years ago
6

Show how the value 0xabcdef12 would be arranged in memory of a little-endian and a big-endian machine. Assume the data is stored

starting at address 0.
Computers and Technology
1 answer:
viva [34]3 years ago
5 0

Answer and Explanation:

For the Big Endian

In a big endian format, the most significant byte is stored at the lowest address and the least significant byte is stored at the highest address. Therefore, the data given is stored as follows:

Address 0 | 1 | 2 | 3

Byte ab | cd | ef | 12

For the Little Endian

In a little endian format, the most significant byte is stored at the highest address (so the least significant byte is stored at the lowest address).

So, the data given would be stored as follows:

Address 0 | 1 | 2 |3

Byte 12 | ef | cd | ab

Hope this Helps!!!

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If a device does not require a driver, it is _____.
xxTIMURxx [149]

Answer:

(B) plug and play

Explanation:

If a device does not require a driver, it is <u>plug and play</u>. Plug and Play (PnP) is a combination of hardware and software support that enables a computer system to recognize and adapt to hardware configuration changes with little or no intervention by a user, including the loading of appropriate drivers.

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4 years ago
What is an example of a source that is less likely to be copyrighted?
Ivahew [28]
An example would be a: motorized vehicle aka a car or a bike etc.
8 0
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What is a special class of adware that collects data about the user and transmits it over the Internet without the user’s knowle
Stells [14]

Answer:

The answer is "Spyware".

Explanation:

Spyware is a supplementary program that is mounted on your customer computer and it transmits the data without authorization via the Internet. Its business includes ads, private data, and improvements in the user interface, which is why it has become recognized as the special adware class that collects and transmits user information over the web without the user's permission.

7 0
4 years ago
Create a test that prompts a user to enter a 5-digit PIN in the main procedure and then calls Validate PIN procedure to validate
Pie

Answer:

See explaination

Explanation:

#include <iomanip>

#include <string>

#include <fstream>

using namespace std;

void getName(string);

void getPin(int,int);

void displayMenu(int);

string name;

int pin1, pin2, ch ;

int main()

{

cout << fixed << showpoint << setprecision(2);

getName(name);

getPin(pin1,pin2);

displayMenu(ch );

system("pause");

}

void getName(string name)

{

cout << "Enter your name: ";

cin >> name;

if (name = "AXBY") || (name != "ABGSHY"))

{

getName(name);

}

}

void getPin(int pin1, int pin2)

{

string name;

if (name == "AXBY")

{

cout << "Please Enter Pin: ";

cin >> pin1;

if (pin1 != 4433)

{

cout << "Pin incorrect please try again!";

cin >> pin1;

}

}

if (name == "ABGSHY")

{

cout << "Please Enter Pin: ";

cin >> pin2;

if (pin2 != 2849)

{

cout << "Pin incorrect please try again!";

cin >> pin2;

}

}

}

void displayMenu(int ch )

{

cout << "[1] Check your account Balance\n"

"[2] Withdraw funds\n"

"[3] Deposit funds\n"

"Please Enter Choice: ";

cin >> ch ;

if (ch != 1 || ch != 2 || ch != 3)

{

cout << "Choice is incorrect, enter choice now: ";

cin >> ch ;

}

}

4 0
4 years ago
The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
poizon [28]

Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

   Random randNum = new Random();

   int people = 5;

   int[] birth_Day = new int[365+1];

   // setting up birthsdays

   for (int i = 0; i < birth_Day.length; i++)

       birth_Day[i] = i + 1;

 

   int iteration;

   // varying number n

   while (people <= 100)

   {

       System.out.println("Number of people: " + people);

       // creating new birth day array

       int[] newbirth_Day = new int[people];

       int count = 0;

       iteration = 100000;

       while(iteration != 0)

       {

           count = 0;

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               // generating random birth day

               int day = randNum.nextInt(365);

               newbirth_Day[i] = birth_Day[day];

           }

           // check for same birthdays

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               int bday = newbirth_Day[i];

               for (int j = i+1; j < newbirth_Day.length; j++)

               {

                   if (bday == newbirth_Day[j])

                   {

                       count++;

                       break;

                   }

               }

           }

           iteration = iteration - 1;

       }

       System.out.println("Probability: " + count + "/" + 100000);

       System.out.println();

       people += 5;

   }

}

}

4 0
4 years ago
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