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Vladimir79 [104]
2 years ago
14

Assume that machines A and B are on the same network 10.3.2.0/24. Machine A sends out spoofed packets, and Machine B tries to sn

iff on the network. When Machine A spoofs packets with a destination 1.2.3.4, Machine B can always observe the spoofed packets. However, when Machine A tries to spoof packets with a destination IP address 10.3.2.30, Machine B cannot see the spoofed packets. There is nothing wrong with the spoofing or sniffing program. Apparently, the spoofed packet has never been sent out. What could be the reason
Computers and Technology
1 answer:
Zanzabum2 years ago
7 0

Answer:

Explanation:

Assuming the spoofed 1.2.3.4, it will be sent out ARP to a router that is alive and get it MAC. Then the spoofed packet will be able to be sent out. If A spoofed 10.0.2.30, because it is a local address, it will send an ARP to the machine for been able to get the MAC. But, the IP address 10.0.2.30 is not original making it to be fake, so it will not replay to A, for this reason the spoofed packet cannot be sent out.

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Mobile computing has two major characteristics that differentiate it from other forms of computing. What are these two character
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mobility and broad reach  

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Write a program to declare a matrix A[][] of order (MXN) where ‘M’ is the number of rows and ‘N’ is the
Liula [17]

Answer:

import java.io.*;

import java.util.Arrays;

class Main {

   public static void main(String args[])

   throws IOException{

       // Set up keyboard input

       InputStreamReader in = new InputStreamReader(System.in);

       BufferedReader br = new BufferedReader(in);

 

       // Prompt for dimensions MxN of the matrix

       System.out.print("M = ");

       int m = Integer.parseInt(br.readLine());

       System.out.print("N = ");

       int n = Integer.parseInt(br.readLine());

       // Check if input is within bounds, exit if not

       if(m <= 2 || m >= 10 || n <= 2 || n >= 10){

           System.out.println("Matrix size out of range.");

           return;

       }

       // Declare the matrix as two-dimensional int array

       int a[][] = new int[m][n];

 

       // Prompt for values of the matrix elements

       System.out.println("Enter elements of matrix:");

       for(int i = 0; i < m; i++){

           for(int j = 0; j < n; j++){

               a[i][j] = Integer.parseInt(br.readLine());

           }

       }

       // Output the original matrix

       System.out.println("Original Matrix:");

       printMatrix(a);

       // Sort each row

       for(int i = 0; i < m; i++){

         Arrays.sort(a[i]);

       }

       // Print sorted matrix

       System.out.println("Matrix after sorting rows:");

       printMatrix(a);

   }

   // Print the matrix elements separated by tabs

   public static void printMatrix(int[][] a) {

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

           for(int j = 0; j < a[i].length; j++)

               System.out.print(a[i][j] + "\t");

           System.out.println();

       }

   }

}

Explanation:

I fixed the mistake in the original code and put comments in to describe each section. The mistake was that the entire matrix was sorted, while only the individual rows needed to be sorted. This even simplifies the program. I also factored out a printMatrix() method because it is used twice.

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