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ANTONII [103]
3 years ago
13

Help it don’t let me click what do I do

Computers and Technology
1 answer:
iogann1982 [59]3 years ago
3 0

Answer: Try resetting you device, or add a mouse to it.

Explanation:

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Match the Job description with the job titles.
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Find the error in the following code fragment. int a: System.out.print(a):​
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3 years ago
Write a program that creates a linked list object of 10 characters and creates a second list object containing a copy of the fir
slamgirl [31]

Answer:

Below code is in c++ with explanation.

Explanation:

Explanation is in comment format, any line starting with //.

#include <iostream>

#include <cstdlib>

using namespace std;

//creating a Node struct which holds node data

struct Node {  

  char data;  

  struct Node *next;  

};  

//creating variable of Node as head

// this head is the start point of ordered Linked list

struct Node* head = NULL;    

// This method is used to insert new value in node

void insert(char new_data) {  

// allocating the memory for new node

  struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));  

// assigning the data to new node

  new_node->data = new_data;  

// If head == null then it means we are inserting first value

// it will return after creating first node and assign it to

// head as head is the entry point

  if(head == NULL){

    new_node->next=NULL;

   head=new_node;

    return;

  }

// Now we need to treverse to end of linked list

// that's why creating a lastNode variable  

// As we don't want our head to point any other node

// except first node

  Node* lastNode = head;

  while(lastNode->next != NULL){

   lastNode=lastNode->next;

  }

// as last node next will be null to end the traversing  

// on linked list

  new_node->next= NULL;

// assigning last node to its previous node  

  lastNode->next = new_node;  

   

}  

// This variable will hold the first node for reverse linklist

struct Node* reverseHead = NULL;    

// This method will reverse the ordered linklist

void reversLinkList(){

  Node* lastNode = head;

// traversing the ordered linklist

  while(lastNode!= NULL){

// creating new linked list

   struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));  

   new_node->data = lastNode->data;  

   new_node->next = reverseHead;  

   reverseHead = new_node;  

   lastNode = lastNode->next;

  }

   

}

// display method will travers the linklist that

// was passed throw paramter

void display(Node* node) {  

  struct Node* ptr;

  ptr = node;

  while (ptr != NULL) {  

     cout<< ptr->data <<" ";  

     ptr = ptr->next;  

  }  

}  

// main method is the entry point of this program

int main() {  

// adding 10 nodes by calling insert method

  insert('a');

  insert('b');

  insert('c');

  insert('d');

  insert('e');

  insert('f');

  insert('g');

  insert('h');

  insert('i');

  insert('j');

  cout<<"The Ordered linked list is: ";

// calling display method with head to print ordered linklist

  display(head);  

  cout<<endl;

  cout<<"The Reversed linked list is: ";

// calling reversLinkList to reverse the ordered linklist

  reversLinkList();

// calling display method with reverseHead to display  

// reversed linklist

  display(reverseHead);

   

  return 0;  

}

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Annette [7]

The <u>TCP</u> hijacking attack uses ip spoofing to enable an attacker to impersonate another entity on the network.

<h3>What is a TCP hijacking attack?</h3>

TCP is a term that connote the Transmission Control Protocol a communications standard that helps application programs as well as computing devices to pass messages over a network.

This hijack is known to be a kind of cyber attack where it is found to be an authorized user that tend to gets control to a legitimate linkage of another client in the network.

Note that by the hijacked of the said TCP/IP session, the attacker can be able to read and change transmitted data packets and thus,The <u>TCP</u> hijacking attack uses ip spoofing to enable an attacker to impersonate another entity on the network.

Learn more about <u>TCP</u> hijacking attack from

brainly.com/question/15052311

#SPJ1

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1 year ago
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