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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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1-(50 points) The function sum_n_avgcomputes the sum and the average of three input arguments and relays its results through two
MAXImum [283]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

double *sum_n_avg(double n1, double n2, double n3);

int main () {

   double n1, n2, n3;

   cout<<"Enter 3 inputs: ";

   cin>>n1>>n2>>n3;

   double *p;

   p = sum_n_avg(n1,n2,n3);

   cout<<"Sum: "<<*(p + 0) << endl;

   cout<<"Average: "<<*(p + 1) << endl;

  return 0;

}

double *sum_n_avg(double n1, double n2, double n3) {

  static double arr[2];

  arr[0] = n1 + n2 + n3;

  arr[1] = arr[0]/3;

  return arr;

}

Explanation:

This defines the function prototype

double *sum_n_avg(double n1, double n2, double n3);

The main begins here

int main () {

This declares variables for input

   double n1, n2, n3;

This prompts the user for 3 inputs

   cout<<"Enter 3 inputs: ";

This gets user inputs

   cin>>n1>>n2>>n3;

This declares a pointer to get the returned values from the function

   double *p;

This passes n1, n2 and n3 to the function and gets the sum and average, in return.

   p = sum_n_avg(n1,n2,n3);

Print sum and average

<em>    cout<<"Sum: "<<*(p + 0) << endl;</em>

<em>    cout<<"Average: "<<*(p + 1) << endl;</em>

  return 0;

}

The function begins here

double *sum_n_avg(double n1, double n2, double n3) {

Declare a static array

  static double arr[2];

Calculate sum

  arr[0] = n1 + n2 + n3;

Calculate average

  arr[1] = arr[0]/3;

Return the array to main

  return arr;

}

5 0
1 year ago
The maximum number of times the decrease key operation performed in Dijkstra's algorithm will be equal to ___________
katovenus [111]

Answer:

b) Single source shortest path

Explanation:

These are the options for the question

a) All pair shortest path

b) Single source shortest path

c) Network flow

d) Sorting

Dijkstra's algorithm is algorithm by

Edsger Dijkstra arround the year 1956, this algorithm determine the shortest path, this path could be between two nodes/vertice of a graph.

The steps involves are;

✓setting up of the distances base on the algorithm.

✓ calculation of the first vertice up to vertice adjacent to it

✓The shortest path result.

It should be noted that maximum number of times the decrease key operation performed in Dijkstra's algorithm will be equal to Single source shortest path.

3 0
1 year ago
A summer camp offers a morning session and an afternoon session.
defon

Answer:

(A) IF (IsFound

(afternoonList, child))

{

APPEND (lunchList, child)

}

Hope this helps!

5 0
2 years ago
Write a function named shout. The function should accept a string argument and display it in uppercase with an exclamation mark
Andrei [34K]

Answer:

void shout(String w) {

System.print.out(w + "!")

}

3 0
1 year ago
c++ You are given an array A representing heights of students. All the students are asked to stand in rows. The students arrive
Lilit [14]

The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.

#include<stdio.h>

#include<string.h>

 int* uniqueValue(int input1,int input2[])

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   int left, current;

   static int arr[4] = {0};

   int i      = 0;

     for(i=0;i<input1;i++)

      {

         current = input2[i];

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         if(current > 0)

         left    = arr[(current-1)];

      if(left == 0 && arr[current] == 0)

       {

       arr[current] = input1-current;

       }

       else

   {

       for(int j=(i+1);j<input1;j++)

       {

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           {

               left = arr[(j-1)];

               arr[j] = left - 1;

           }

       }

   }

}

return arr;

}

4 0
2 years ago
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