The document discusses implementing various operations for a linked list data structure including: constructors, destructors, insertion, removal, traversal, and more. It provides template code definitions for a ListNode class and List class to represent nodes and the overall linked list. It also includes functions to insert and delete nodes at different positions, find the size, reverse, merge, and check for cycles in linked lists.
10- Consider the following data structure used for implementing a link.docxtodd991
10. Consider the following data structure used for implementing a linked list: struct Node int data. Node *next; Implement the function that takes a pointer to the first node of a linked list and an integer value as arguments. It returns -1 if the input value is not equal to the data value of any item in the list. Otherwise, it returns the location of an item in the list whose data value is equal to the input value. Assume that the location of the first item in the list is 0. int search List (Node *head, int value) write the function body
Solution
#include <iostream>
using namespace std;
struct Node{
int data;
Node *next;
};
//inserting element at front side in list
void insert(Node **head,int data){
struct Node *temp;
if(*head==NULL){
(*head)=(struct Node*)malloc(sizeof(struct Node));
(*head)->data=data;
(*head)->next=NULL;
}else{
temp=(struct Node*)malloc(sizeof(struct Node));
temp->data=data;
temp->next=(*head);
(*head)=temp;
}
}
// searching element in list
int searchList(Node *head, int value){
int i=0;int flag=0;
while(head!=NULL){
if(head->data==value){
flag=1;
return i;
}
i++;
head=head->next;
}
if(flag==0)
return -1;
}
void printElement(Node *head){
while(head!=NULL){
cout<<head->data;
head=head->next;
}
}
int main()
{
struct Node *head=NULL;
insert(&head,10);
insert(&head,20);
insert(&head,30);
insert(&head,40);
insert(&head,50);
//printElement(head);
int rValue=searchList(head,30);
if(rValue==-1){
cout<<\"Value do not exist in list\"<<endl;
}
else{
cout<<\"Value exist at position \"<<rValue<<endl;
}
return 0;
}
/*sample output*/
searchList(head,30);
Value exist at 2
searchList(head,60);
Value do not exist in list
.
This presentations gives an introduction to the data structure linked-lists. I discuss the implementation of header-based linked-lists in C. The presentation runs through the code and provides the visualization of the code w.r.t pointers.
Hi,I have added the methods and main class as per your requirement.pdfannaelctronics
Hi,
I have added the methods and main class as per your requirement. it is working fine now.
Highlighted the code changes.
MyLinkedListTest.java
public class MyLinkedListTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
MyLinkedList list = new MyLinkedList();
list.addFirst(\"AAAAA\");
list.add(\"1\");
list.add(\"2\");
list.add(\"3\");
list.addLast(\"ZZZZZ\");
System.out.println(\"Linked List Elements : \"+list.toString());
System.out.println(\"First element: \"+list.getFirst());
System.out.println(\"Second element: \"+list.getLast());
System.out.println(\"Removed First Element: \"+list.removeFirst());
System.out.println(\"Now Linked List Elements : \"+list.toString());
System.out.println(\"Removed Last Element: \"+list.removeLast());
System.out.println(\"Now Linked List Elements : \"+list.toString());
System.out.println(\"Removed all element: \");
list.removeAll();
System.out.println(\"Now Linked List Elements : \"+list.toString());
}
}
MyLinkedList.java
public class MyLinkedList extends MyAbstractList {
private Node head, tail;
/** Create a default list */
public MyLinkedList() {
}
/** Create a list from an array of objects */
public MyLinkedList(E[] objects) {
super(objects);
}
/** Return the head element in the list */
public E getFirst() {
if (size == 0) {
return null;
}
else {
return (E)head.element;
}
}
/** Return the last element in the list */
public E getLast() {
if (size == 0) {
return null;
}
else {
return (E)tail.element;
}
}
/** Add an element to the beginning of the list */
public void addFirst(E e) {
Node newNode = new Node(e); // Create a new node
newNode.next = head; // link the new node with the head
head = newNode; // head points to the new node
size++; // Increase list size
if (tail == null) // the new node is the only node in list
tail = head;
}
/** Add an element to the end of the list */
public void addLast(E e) {
Node newNode = new Node(e); // Create a new for element e
if (tail == null) {
head = tail = newNode; // The new node is the only node in list
}
else {
tail.next = newNode; // Link the new with the last node
tail = tail.next; // tail now points to the last node
}
size++; // Increase size
}
/** Add a new element at the specified index in this list
* The index of the head element is 0 */
public void add(int index, E e) {
if (index == 0) {
addFirst(e);
}
else if (index >= size) {
addLast(e);
}
else {
Node current = head;
for (int i = 1; i < index; i++) {
current = current.next;
}
Node temp = current.next;
current.next = new Node(e);
(current.next).next = temp;
size++;
}
}
/** Remove the head node and
* return the object that is contained in the removed node. */
public E removeFirst() {
if (size == 0) {
return null;
}
else {
Node temp = head;
head = head.next;
size--;
if (head == null) {
tail = null;
}
return (E)temp.element;
}
}
/** Remove all elements from list */
public void removeAll() {
while(true) {
if (size == 0) {
break;
}
else {
Node temp = head;
head =.
10- Consider the following data structure used for implementing a link.docxtodd991
10. Consider the following data structure used for implementing a linked list: struct Node int data. Node *next; Implement the function that takes a pointer to the first node of a linked list and an integer value as arguments. It returns -1 if the input value is not equal to the data value of any item in the list. Otherwise, it returns the location of an item in the list whose data value is equal to the input value. Assume that the location of the first item in the list is 0. int search List (Node *head, int value) write the function body
Solution
#include <iostream>
using namespace std;
struct Node{
int data;
Node *next;
};
//inserting element at front side in list
void insert(Node **head,int data){
struct Node *temp;
if(*head==NULL){
(*head)=(struct Node*)malloc(sizeof(struct Node));
(*head)->data=data;
(*head)->next=NULL;
}else{
temp=(struct Node*)malloc(sizeof(struct Node));
temp->data=data;
temp->next=(*head);
(*head)=temp;
}
}
// searching element in list
int searchList(Node *head, int value){
int i=0;int flag=0;
while(head!=NULL){
if(head->data==value){
flag=1;
return i;
}
i++;
head=head->next;
}
if(flag==0)
return -1;
}
void printElement(Node *head){
while(head!=NULL){
cout<<head->data;
head=head->next;
}
}
int main()
{
struct Node *head=NULL;
insert(&head,10);
insert(&head,20);
insert(&head,30);
insert(&head,40);
insert(&head,50);
//printElement(head);
int rValue=searchList(head,30);
if(rValue==-1){
cout<<\"Value do not exist in list\"<<endl;
}
else{
cout<<\"Value exist at position \"<<rValue<<endl;
}
return 0;
}
/*sample output*/
searchList(head,30);
Value exist at 2
searchList(head,60);
Value do not exist in list
.
This presentations gives an introduction to the data structure linked-lists. I discuss the implementation of header-based linked-lists in C. The presentation runs through the code and provides the visualization of the code w.r.t pointers.
Hi,I have added the methods and main class as per your requirement.pdfannaelctronics
Hi,
I have added the methods and main class as per your requirement. it is working fine now.
Highlighted the code changes.
MyLinkedListTest.java
public class MyLinkedListTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
MyLinkedList list = new MyLinkedList();
list.addFirst(\"AAAAA\");
list.add(\"1\");
list.add(\"2\");
list.add(\"3\");
list.addLast(\"ZZZZZ\");
System.out.println(\"Linked List Elements : \"+list.toString());
System.out.println(\"First element: \"+list.getFirst());
System.out.println(\"Second element: \"+list.getLast());
System.out.println(\"Removed First Element: \"+list.removeFirst());
System.out.println(\"Now Linked List Elements : \"+list.toString());
System.out.println(\"Removed Last Element: \"+list.removeLast());
System.out.println(\"Now Linked List Elements : \"+list.toString());
System.out.println(\"Removed all element: \");
list.removeAll();
System.out.println(\"Now Linked List Elements : \"+list.toString());
}
}
MyLinkedList.java
public class MyLinkedList extends MyAbstractList {
private Node head, tail;
/** Create a default list */
public MyLinkedList() {
}
/** Create a list from an array of objects */
public MyLinkedList(E[] objects) {
super(objects);
}
/** Return the head element in the list */
public E getFirst() {
if (size == 0) {
return null;
}
else {
return (E)head.element;
}
}
/** Return the last element in the list */
public E getLast() {
if (size == 0) {
return null;
}
else {
return (E)tail.element;
}
}
/** Add an element to the beginning of the list */
public void addFirst(E e) {
Node newNode = new Node(e); // Create a new node
newNode.next = head; // link the new node with the head
head = newNode; // head points to the new node
size++; // Increase list size
if (tail == null) // the new node is the only node in list
tail = head;
}
/** Add an element to the end of the list */
public void addLast(E e) {
Node newNode = new Node(e); // Create a new for element e
if (tail == null) {
head = tail = newNode; // The new node is the only node in list
}
else {
tail.next = newNode; // Link the new with the last node
tail = tail.next; // tail now points to the last node
}
size++; // Increase size
}
/** Add a new element at the specified index in this list
* The index of the head element is 0 */
public void add(int index, E e) {
if (index == 0) {
addFirst(e);
}
else if (index >= size) {
addLast(e);
}
else {
Node current = head;
for (int i = 1; i < index; i++) {
current = current.next;
}
Node temp = current.next;
current.next = new Node(e);
(current.next).next = temp;
size++;
}
}
/** Remove the head node and
* return the object that is contained in the removed node. */
public E removeFirst() {
if (size == 0) {
return null;
}
else {
Node temp = head;
head = head.next;
size--;
if (head == null) {
tail = null;
}
return (E)temp.element;
}
}
/** Remove all elements from list */
public void removeAll() {
while(true) {
if (size == 0) {
break;
}
else {
Node temp = head;
head =.
Please need help on following program using c++ language. Please inc.pdfnitinarora01
Please need help on following program using c++ language. Please include all header files and
main file with their own title.
Extend the class linkedListType by adding the following operations:
Write a function that returns the info of the kth element of the linked list. If no such element
exists, terminate the program.
Write a function that deletes the kth element of the linked list. If no such element exists,
terminate the program.
Provide the definitions of these functions in the class linkedListType. Also, write a program to
test these functions. (Use either the class unorderedLinkedList or the
class orderedLinkedList to test your function.)
Solution
C++ Code for the given problem is below.....
.....................................................
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
//using namespace std;
/*
* Node Declaration
*/
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
/*
* Class Declaration
*/
class linkedListType
{
public:
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
int item,ch,kth;
node *ptr,*head,*temp;
};
/*
* Main :contains menu
*/
void main()
{
int item,ch,kth;
node *ptr,*head,*temp;
linkedListType s1;
clrscr();
while (1)
{
while(1)
{
cout<<\"PRESS (1) FOR INSERT INTO LIST\"<>ch;
switch(ch)
{
case 1:
cout<<\"ENTER THE VALUE \"<>item;
s1.insfrombeg(&head,item);
break;
case 2:
s1.traverse(head);
break;
case 3:
cout<<\"ENTER THE Kth Element which you want\'s to delete\"<>kth;
s1.delfromkth(&head,kth);
break;
case 4:
cout<<\"ENTER THE Kth Element which you want\'s to search \ \"<>kth;
s1.findthekthnode(head,kth);
break;
case 5:
exit(1);
}
}
// getch();
}
}
/*
* Inserting element in beginning
*/
void linkedListType::insfrombeg(node **head,int item)
{
temp= ( node * ) malloc (sizeof( node ));
temp->info= item;
temp->next= NULL;
if(*head == NULL)
{
temp->next= *head;
*head= temp;
}
else
{
temp->next= *head;
*head= temp;
}
}
/*
* Delete element at a given position
*/
void linkedListType::delfromkth(node **head,int kth)
{ node* temp1=*head,*temp2;
if(*head ==NULL)
{
cout<<\"LINK LIST IS EMPTY\ \"<next;
free(temp);
}
else
for(int j=0;jnext;
temp2= temp1->next;
temp1->next=temp2->next;
free(temp2);
}
/*
* Searching an element
*/
void linkedListType::findthekthnode(node *head,int kth)
{
ptr=head;
int i=1;
while((ptr!=NULL))
{
if(i==kth)
{
cout<<\"\ The Element at\"<< kth<< \"position with value is \\t
\"<info<info<next;
i++;
}
}
}
/*
* Traverse Link List
*/
void linkedListType::traverse(node *head)
{
ptr = head;
while((ptr->next)->next!=NULL)
{
cout<info<<\"\\t\";
ptr=ptr->next;
}
cout<
#include
#include
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
int item,ch,kth,total=1;
node *ptr,*head,*temp;
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
void main()
{
clrscr();
while(1)
{
printf(\"PRESS .
To complete the task, you need to fill in the missing code. I’ve inc.pdfezycolours78
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
–PLS write program in c++Recursive Linked List OperationsWrite a.pdfpasqualealvarez467
–PLS write program in c++
Recursive Linked List Operations
Write a C++ to manage a list of string elements. Besides the basic operations expected of a
linked list (add an element, remove an element, check for empty list, report size (number) of
elements in the list, and print out the content of the list), the linked list class also needs to have a
method called sort() and a method called reverse() the manipulate the string elements. The sort
method rearranges the elements in the list so they are sorted in alphabetical order. The reverse
method reverses the order of the elements of the list. The class should use recursion to
implement the sort and reverse operations
Solution
#include
#include
#include
#include
struct node //declaring node
{
int info;
struct node *next;
}*start;
class single_llist //declaring class
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
main()
{
int choice, nodes, position, i;
char str[100];
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (strcmp(ptr->info > s->info))
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
void single_llist::delete_pos() //delete element from given position
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2.
Write a program that accepts an arithmetic expression of unsigned in.pdfJUSTSTYLISH3B2MOHALI
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
take the following code and give details of what each line of code i.pdffastechsrv
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
ItemNodeh include ltiostreamgt include ltstring.pdfacmefit
ItemNode.h:
#include <iostream>
#include <string>
using namespace std;
class ItemNode {
private:
string item;
ItemNode* nextNodeRef;
public:
// Constructor
ItemNode() {
item = "";
nextNodeRef = NULL;
}
// Constructor
ItemNode(string itemInit) {
this->item = itemInit;
this->nextNodeRef = NULL;
}
// Constructor
ItemNode(string itemInit, ItemNode *nextLoc) {
this->item = itemInit;
this->nextNodeRef = nextLoc;
}
// Insert node after this node.
void InsertAfter(ItemNode &nodeLoc) {
ItemNode* tmpNext;
tmpNext = this->nextNodeRef;
this->nextNodeRef = &nodeLoc;
nodeLoc.nextNodeRef = tmpNext;
}
// Define InsertAtEnd() function that inserts a node
// to the end of the linked list
void InsertAtEnd(string item) {
ItemNode* newNode = new ItemNode(item);
if (nextNodeRef == NULL) {
nextNodeRef = newNode;
} else {
ItemNode* current = nextNodeRef;
while (current->nextNodeRef != NULL) {
current = current->nextNodeRef;
}
current->nextNodeRef = newNode;
}
}
// Get location pointed by nextNodeRef
ItemNode* GetNext() {
return this->nextNodeRef;
}
void PrintNodeData() {
cout << this->item << endl;
}
};
int main() {
int numItems;
cin >> numItems;
ItemNode head;
ItemNode* current = &head;
for (int i = 0; i < numItems; i++) {
string item;
cin >> item;
current->InsertAtEnd(item);
current = current->GetNext();
}
current = head.GetNext();
while (current != NULL) {
current->PrintNodeData();
current = current->GetNext();
}
return 0;
}
main.cpp (only for viewing)
#include "ItemNode.h"
int main() {
ItemNode *headNode; // Create intNode objects
ItemNode *currNode;
ItemNode *lastNode;
string item;
int i;
int input;
// Front of nodes list
headNode = new ItemNode();
lastNode = headNode;
cin >> input;
for (i = 0; i < input; i++) {
cin >> item;
currNode = new ItemNode(item);
lastNode->InsertAtEnd(currNode);
lastNode = currNode;
}
// Print linked list
currNode = headNode->GetNext();
while (currNode != NULL) {
currNode->PrintNodeData();
currNode = currNode->GetNext();
}
}
in c++ please. Thank you!
18.18 LAB: Grocery shopping list (linked list: inserting at the end of a list) Given main0, define an
InsertAtEnd() member function in the ItemNode class that adds an element to the end of a linked
list. DO NOT print the dummy head node. Ex. if the input is: begin{tabular}{|l} hline 4 Kale Lettuce
Carrots Peanuts end{tabular} where 4 is the number of items to be inserted; Kale, Lettuce,
Carrots, Peanuts are the names of the items to be added at the end of the list. The output is:.
You are asked with writing a program in C that manages contact infor.pdfFOREVERPRODUCTCHD
You are asked with writing a program in C that manages contact information for a group of
people. The program should save the first name, last name, and telephone number for up to 12
people. The program should have options to add a person, delete a person, update the
information for a person, and display all information for all current entries. How could you go
about breaking up the programming work into a set of subproblems that could be implemented
separately?
Solution
Whenever we asked about these type of problems.we just go to data structures.i.e,Data structures
that are suitable to our problem whether it may be linked lists,arrays,trees.Here we are taking
about a person information that to it is not about a single person information.
1)Identify data structure.Here i think linked list data structure can be used because array cannot
store all information about a person.and arrays cannot support delete operation like that.
2)1st sub problem:creating a structure to a single person
3)2nd sub problem:add a person:inserting data in linked list by creating a single node
3)updating information:modifying node information in single linked list
4)delete the information of a person:delete a single node information in single linked list
5)display all information for all current entries:traversing a single linked list.
each step is considered to subproblem or function in \'C\' language..
Write a short paragraph explaining what the exploit does in Assembly.pdfFOREVERPRODUCTCHD
Write a short paragraph explaining what the exploit does in Assembly , and how it works. How
could the
programmer of vecho have prevented this from happening?
Solution
Exploit in assembly allows us to gain unauthorized root priviledge by using buffer overflow. It
allows hackers to gain access to addresses beyond the eip register. Exploits are generally used in
developing virus. The command vecho, which is similar to the echo command is vulnerable and
used in exploits.
The programmers of vecho should have checked for the address in the eip register and any
attempt to access addresses beyond it should have thrown an error..
More Related Content
Similar to implement the ListLinked ADT (the declaration is given in ListLinked.pdf
Please need help on following program using c++ language. Please inc.pdfnitinarora01
Please need help on following program using c++ language. Please include all header files and
main file with their own title.
Extend the class linkedListType by adding the following operations:
Write a function that returns the info of the kth element of the linked list. If no such element
exists, terminate the program.
Write a function that deletes the kth element of the linked list. If no such element exists,
terminate the program.
Provide the definitions of these functions in the class linkedListType. Also, write a program to
test these functions. (Use either the class unorderedLinkedList or the
class orderedLinkedList to test your function.)
Solution
C++ Code for the given problem is below.....
.....................................................
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
//using namespace std;
/*
* Node Declaration
*/
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
/*
* Class Declaration
*/
class linkedListType
{
public:
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
int item,ch,kth;
node *ptr,*head,*temp;
};
/*
* Main :contains menu
*/
void main()
{
int item,ch,kth;
node *ptr,*head,*temp;
linkedListType s1;
clrscr();
while (1)
{
while(1)
{
cout<<\"PRESS (1) FOR INSERT INTO LIST\"<>ch;
switch(ch)
{
case 1:
cout<<\"ENTER THE VALUE \"<>item;
s1.insfrombeg(&head,item);
break;
case 2:
s1.traverse(head);
break;
case 3:
cout<<\"ENTER THE Kth Element which you want\'s to delete\"<>kth;
s1.delfromkth(&head,kth);
break;
case 4:
cout<<\"ENTER THE Kth Element which you want\'s to search \ \"<>kth;
s1.findthekthnode(head,kth);
break;
case 5:
exit(1);
}
}
// getch();
}
}
/*
* Inserting element in beginning
*/
void linkedListType::insfrombeg(node **head,int item)
{
temp= ( node * ) malloc (sizeof( node ));
temp->info= item;
temp->next= NULL;
if(*head == NULL)
{
temp->next= *head;
*head= temp;
}
else
{
temp->next= *head;
*head= temp;
}
}
/*
* Delete element at a given position
*/
void linkedListType::delfromkth(node **head,int kth)
{ node* temp1=*head,*temp2;
if(*head ==NULL)
{
cout<<\"LINK LIST IS EMPTY\ \"<next;
free(temp);
}
else
for(int j=0;jnext;
temp2= temp1->next;
temp1->next=temp2->next;
free(temp2);
}
/*
* Searching an element
*/
void linkedListType::findthekthnode(node *head,int kth)
{
ptr=head;
int i=1;
while((ptr!=NULL))
{
if(i==kth)
{
cout<<\"\ The Element at\"<< kth<< \"position with value is \\t
\"<info<info<next;
i++;
}
}
}
/*
* Traverse Link List
*/
void linkedListType::traverse(node *head)
{
ptr = head;
while((ptr->next)->next!=NULL)
{
cout<info<<\"\\t\";
ptr=ptr->next;
}
cout<
#include
#include
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
int item,ch,kth,total=1;
node *ptr,*head,*temp;
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
void main()
{
clrscr();
while(1)
{
printf(\"PRESS .
To complete the task, you need to fill in the missing code. I’ve inc.pdfezycolours78
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
–PLS write program in c++Recursive Linked List OperationsWrite a.pdfpasqualealvarez467
–PLS write program in c++
Recursive Linked List Operations
Write a C++ to manage a list of string elements. Besides the basic operations expected of a
linked list (add an element, remove an element, check for empty list, report size (number) of
elements in the list, and print out the content of the list), the linked list class also needs to have a
method called sort() and a method called reverse() the manipulate the string elements. The sort
method rearranges the elements in the list so they are sorted in alphabetical order. The reverse
method reverses the order of the elements of the list. The class should use recursion to
implement the sort and reverse operations
Solution
#include
#include
#include
#include
struct node //declaring node
{
int info;
struct node *next;
}*start;
class single_llist //declaring class
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
main()
{
int choice, nodes, position, i;
char str[100];
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (strcmp(ptr->info > s->info))
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
void single_llist::delete_pos() //delete element from given position
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2.
Write a program that accepts an arithmetic expression of unsigned in.pdfJUSTSTYLISH3B2MOHALI
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
take the following code and give details of what each line of code i.pdffastechsrv
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
ItemNodeh include ltiostreamgt include ltstring.pdfacmefit
ItemNode.h:
#include <iostream>
#include <string>
using namespace std;
class ItemNode {
private:
string item;
ItemNode* nextNodeRef;
public:
// Constructor
ItemNode() {
item = "";
nextNodeRef = NULL;
}
// Constructor
ItemNode(string itemInit) {
this->item = itemInit;
this->nextNodeRef = NULL;
}
// Constructor
ItemNode(string itemInit, ItemNode *nextLoc) {
this->item = itemInit;
this->nextNodeRef = nextLoc;
}
// Insert node after this node.
void InsertAfter(ItemNode &nodeLoc) {
ItemNode* tmpNext;
tmpNext = this->nextNodeRef;
this->nextNodeRef = &nodeLoc;
nodeLoc.nextNodeRef = tmpNext;
}
// Define InsertAtEnd() function that inserts a node
// to the end of the linked list
void InsertAtEnd(string item) {
ItemNode* newNode = new ItemNode(item);
if (nextNodeRef == NULL) {
nextNodeRef = newNode;
} else {
ItemNode* current = nextNodeRef;
while (current->nextNodeRef != NULL) {
current = current->nextNodeRef;
}
current->nextNodeRef = newNode;
}
}
// Get location pointed by nextNodeRef
ItemNode* GetNext() {
return this->nextNodeRef;
}
void PrintNodeData() {
cout << this->item << endl;
}
};
int main() {
int numItems;
cin >> numItems;
ItemNode head;
ItemNode* current = &head;
for (int i = 0; i < numItems; i++) {
string item;
cin >> item;
current->InsertAtEnd(item);
current = current->GetNext();
}
current = head.GetNext();
while (current != NULL) {
current->PrintNodeData();
current = current->GetNext();
}
return 0;
}
main.cpp (only for viewing)
#include "ItemNode.h"
int main() {
ItemNode *headNode; // Create intNode objects
ItemNode *currNode;
ItemNode *lastNode;
string item;
int i;
int input;
// Front of nodes list
headNode = new ItemNode();
lastNode = headNode;
cin >> input;
for (i = 0; i < input; i++) {
cin >> item;
currNode = new ItemNode(item);
lastNode->InsertAtEnd(currNode);
lastNode = currNode;
}
// Print linked list
currNode = headNode->GetNext();
while (currNode != NULL) {
currNode->PrintNodeData();
currNode = currNode->GetNext();
}
}
in c++ please. Thank you!
18.18 LAB: Grocery shopping list (linked list: inserting at the end of a list) Given main0, define an
InsertAtEnd() member function in the ItemNode class that adds an element to the end of a linked
list. DO NOT print the dummy head node. Ex. if the input is: begin{tabular}{|l} hline 4 Kale Lettuce
Carrots Peanuts end{tabular} where 4 is the number of items to be inserted; Kale, Lettuce,
Carrots, Peanuts are the names of the items to be added at the end of the list. The output is:.
You are asked with writing a program in C that manages contact infor.pdfFOREVERPRODUCTCHD
You are asked with writing a program in C that manages contact information for a group of
people. The program should save the first name, last name, and telephone number for up to 12
people. The program should have options to add a person, delete a person, update the
information for a person, and display all information for all current entries. How could you go
about breaking up the programming work into a set of subproblems that could be implemented
separately?
Solution
Whenever we asked about these type of problems.we just go to data structures.i.e,Data structures
that are suitable to our problem whether it may be linked lists,arrays,trees.Here we are taking
about a person information that to it is not about a single person information.
1)Identify data structure.Here i think linked list data structure can be used because array cannot
store all information about a person.and arrays cannot support delete operation like that.
2)1st sub problem:creating a structure to a single person
3)2nd sub problem:add a person:inserting data in linked list by creating a single node
3)updating information:modifying node information in single linked list
4)delete the information of a person:delete a single node information in single linked list
5)display all information for all current entries:traversing a single linked list.
each step is considered to subproblem or function in \'C\' language..
Write a short paragraph explaining what the exploit does in Assembly.pdfFOREVERPRODUCTCHD
Write a short paragraph explaining what the exploit does in Assembly , and how it works. How
could the
programmer of vecho have prevented this from happening?
Solution
Exploit in assembly allows us to gain unauthorized root priviledge by using buffer overflow. It
allows hackers to gain access to addresses beyond the eip register. Exploits are generally used in
developing virus. The command vecho, which is similar to the echo command is vulnerable and
used in exploits.
The programmers of vecho should have checked for the address in the eip register and any
attempt to access addresses beyond it should have thrown an error..
Write a program to convert a given INFIX into POSTFIX. Make sure .pdfFOREVERPRODUCTCHD
Write a program to convert a given INFIX into POSTFIX. Make sure the program checks for \"(
)\", and check the balances of parentheses.
Write a program to convert a given INFIX into POSTFIX. Make sure the program checks for \"(
)\", and check the balances of parentheses.
Solution
#include
#include
#include
#define size 100
using namespace std;
char s[size];
int top=-1;
push(char z)
{ /*PUSH Function*/
s[++top]=z;
}
char pop()
{ /*POP Function*/
return(s[top--]);
}
int pr(char z)
{
switch(z)
{
case \'#\': return 0;
case \'(\': return 1;
case \'+\':
case \'-\': return 2;
case \'*\':
case \'/\': return 3;
}
}
int main()
{
char in[100],post[100],c,z;
int i=0,k=0;
printf(\"\ \ Enter your Infix Expression : \");
scanf(\"%s\",in);
push(\'#\');
while( (c=in[i++]) != \'\\0\')
{
if( c == \'(\') push(c);
else
if(isalnum(c)) post[k++]=c;
else
if( c == \')\')
{
while( s[top] != \'(\')
post[k++]=pop();
z=pop(); /* Remove ( */
}
else
{ /* Operator */
while( pr(s[top]) >= pr(c) )
post[k++]=pop();
push(c);
}
}
while( s[top] != \'#\') /* Popping from stack */
post[k++]=pop();
post[k]=\'\\0\';
printf(\"\ \ Infix Expression: %s \ \ Postfix Expression: %s\ \",in,post);
}.
What quality control mechanisms should major accounting firms have i.pdfFOREVERPRODUCTCHD
What quality control mechanisms should major accounting firms have in place to ensure that
audit partners have the proper training and experience to supervise audit engagements?
Solution
Quality control measures the major Accounting firm have in place in order to ensure that
auditors have proper training and experience to supervise audit engagements are
(1) Integrity, objectivity and independence: The auditor should be straight forward, honest and
sincere in his approach to his professional work. He should maintain an impartial
attitude and both be and appear to be free of any interest which might be regarded, whatever is
actual effect, as being incompatible with integrity and objectivity.
(2) Confidentiality: The auditor should respect the confidentiality of information acquired in the
course of his work and should not disclose any such information to a third party without The
specific authority or unless there is a legal or professional duty to disclose.
(3) Skills and Competence: The audit should be performed and the report prepared with due
professional care by persons who have adequate training, experience and competence in
auditing. The auditor requires specialised skills and competence along with a continuing
awareness of developments including pronouncements of the accounting body on accounting and
auditing matters, and relevant regulations and statutory requirements.
(4) Work performed by others: When the auditor delegates work to assistants or uses work
performed by other auditors and experts, he continues to be responsible for forming and
expressing his opinion on the financial information.
These are the some of the measures to have proper training and experience to supervise audit
engagements
I WOULD LIKE TO SHARE YOU ABOUT LETTER OF ENGAGEMENT IN SHORT
Letter of Engagement: The legal requirement to get the accounts audited so far extends only to
companies, co-operative societies, and registered societies. In these cases, the respective law
governs the appointment of auditors and their duties. In all other cases, it is a matter of contract.
Thank you hope you might feel it helpfull.
What are three basic ethical principles for journalism Why are ethi.pdfFOREVERPRODUCTCHD
What are three basic ethical principles for journalism? Why are ethics so important for a writer?
Solution
Three major ethical principles are
Truth and Accuracy
Journalists cannot always guarantee ‘truth’, but getting the facts right is the cardinal principle of
journalism. We should always strive for accuracy, give all the relevant facts we have and ensure
that they have been checked. When we cannot corroborate information we should say so.
Independence
Journalists must be independent voices; we should not act, formally or informally, on behalf of
special interests whether political, corporate or cultural. We should declare to our editors – or the
audience – any of our political affiliations, financial arrangements or other personal information
that might constitute a conflict of interest.
Fairness and Impartiality
Most stories have at least two sides. While there is no obligation to present every side in every
piece, stories should be balanced and add context. Objectivity is not always possible, and may
not always be desirable (in the face for example of brutality or inhumanity), but impartial
reporting builds trust and confidence..
What are some of the different versions of UNIX® Why is it importan.pdfFOREVERPRODUCTCHD
What are some of the different versions of UNIX®? Why is it important to have different
versions of UNIX®?
Solution
Versions of UNIX
Many versions of UNIX, including System V Release 4, merge earlier AT&T releases with BSD
features. In last few years ago, there were two main versions:
[1] The line of UNIX releases that started at AT&T (the latest is System V Release 4)
[2] The University of California at Berkeley (the latest version is BSD 4.4).
Some more commercial versions include Solaris, SCO UNIX , SunOS, AIX, HP/UX, and
ULTRIX. The recent POSIX standard for UNIX-like operating systems defines a single interface
to UNIX..
Type in your own words In details, discuss the following questions.pdfFOREVERPRODUCTCHD
Type in your own words
In details, discuss the following questions in three to four pages: 1- Give one of your own
experiences in which your intrinsic motivation changed into extrinsic [LO31| 10 Marks]
motivation as through competition, or rewards. What are some activities you do because you get
external rewards, or because you are avoiding some punishment? 2- LO41[10 Marks]
Solution
Extrinsic motivation refers to behavior that is driven by external rewards such as money, fame,
grades, and praise. This type of motivation arises from outside the individual, as opposed to
intrinsic motivation, which originates inside of the individual.”
? Intrinsic motivation refers to behavior that is driven by internal rewards. In other words, the
motivation to engage in a behavior arises from within the individual because it is intrinsically
rewarding.”
Example of Coversion of intricsic motivation into extrinsic motivation
My personel life experiance is to become CA ....... i want to become CA because i m intrested in
accountancy field from start of 10th class when i solve one problem based on account field very
easily and with intrest taking in solving that problem .... then i decided to take commerce an
choose CA field ..when i came in this field then discussion with the person around me were CA
.....then i knwe All about their struggle and theri popularity after become CA .....that all thing
also motivates us .
We think that is disputable, and in some way, outrageous proposal to withhold reward in school
setting that is based in the studies of motivation involved in Deci\' s and his associates meta-
analysis. From the 70\'s and onwards there was a significant endeavor in the research agendas.
Behavioral model of learning through rigid control of external reward is no longer the dominant
paradigm. Researcher\'s attention has been placed on the contribution of valuable learning that
comes from personal effort and \"inner senses\". Therefore, maybe it is time to move the agenda
to the orientation of the synthesis of external and internal motivation that will engage students,
especially adolescents, within the accepted teaching skills and competencies necessary for
creative and productive future. And we should not forget the fact that in most cases the
researchers were successful in identifying factors that reduce interest than factors that reinforce
them. Only recently, researchers have begun to portray interventions that may enhance the
interest and therefore the value, importance and meaning of the activities or polishing of the
structure of
Activites done for reward.
Twice a first number decreased by a second number is 11. The first nu.pdfFOREVERPRODUCTCHD
Twice a first number decreased by a second number is 11. The first number increased by twice
the second number is 18. Find the numbers. The numbers are (Use a comma to separate answers
as needed.)
Solution
Let x be the first number and
y be the second number
2x-y=11
x+2y=18
4x-2y=22
x+2y=18
Adding abive equations
5x=40
x=8
8+2y=18
y=5.
the largest drum ever constructed was played at the Rotal festival H.pdfFOREVERPRODUCTCHD
the largest drum ever constructed was played at the Rotal festival Hall in london in 1987. it had a
diameter of 13 ft . what was the area of the circular face of the drum?
Solution
diameter of the drum = 13ft
area of circular face of drum = upper circular area + lower circular area
circular area = pi*r^2
r = 13/2 = 6.5 ft
therefore, area =2* pi* 6.5^2 = 265.33 square feet.
Suppose that the material that you are recrystallizing fails to perc.pdfFOREVERPRODUCTCHD
Suppose that the material that you are recrystallizing fails to percipiate out of the cold solvent.
What would you do to recover the material from the solution? Experiment 4: Synthesis of
Salicylic Acid
Solution
You can get back your product, by concentrating the solvent, by evaporating the solvent under
vacuum, at temperature around 40 degreed celcius in rota evaporator..
Summarize the purpose of a WAN and define what makes up a WAN connec.pdfFOREVERPRODUCTCHD
Summarize the purpose of a WAN and define what makes up a WAN connection.
Detail the types of devices used to establish and maintain a WAN connection and list some of the
more common WAN connection methods.
Discuss the differences between WAN and LAN connections, and who is responsible for
providing WAN connections.
Discuss how a modem handles digital information over analog signals.
Describe how to configure and setup remote access protocols.
What factors need careful consideration?
Solution
Remote Access Protocols
Remote access protocols square measure accustomed management the institution of connections
and therefore the transmission of information over WAN links. 3 varieties of remote access
protocols square measure supported by Windows Server 2003 remote access:
Point-to-Point Protocol (PPP): AN industry-standard set of protocols providing the most
effective security, multi-protocol support and ability.
Serial Line net Protocol (SLIP): employed by older remote access servers. A Windows Server
2003 RAS server doesn\'t support SLIP dial-up connections.
Microsoft remote access protocol: additionally called Asynchronous NetBEUI (AsyBEUI) and is
employed by inheritance remote access purchasers, like Windows NT three.1, Windows for
Workgroups, DOS and computer network Manager purchasers.
LAN may be a electronic network that connects computers in tiny areas. WAN may be a network
that covers a broad space victimization personal or public network transports.
The terms {lan|local square measurea network|LAN|computer network} and WAN are usually
confusing for those that aren’t that school savvy. These square measure each connections that
permit users to attach their laptop to a network, as well as the web. computer network is brief for
native space Network, whereas WAN is brief for Wide space Network. These 2 disagree from
one another in distinct ways that. computer network may be a electronic network that connects
computers in tiny areas like home, office, school, corporation, etc. employing a network media.
it\'s helpful for sharing resources like printers, files, games, etc. A computer network network
includes a few of laptop systems connected to every alternative, with one system connected to a
router, electronic equipment or AN outlet for net access. The computer network network is made
victimization cheap technologies like local area network cables, network adapters and hubs.
However, alternative wireless technologies are obtainable to attach the pc through a wireless
access. so as to put together a computer network network, an individual might also need
specialised software system software system. the foremost widespread software system includes
the Microsoft Windows’ net affiliation Sharing (ICS), that permits users to make computer
network. distinction between computer network and WAN
Key Difference: computer network may be a electronic network that connects computers in tiny
areas. WAN may be a network that covers a broad space victimizat.
SOS Please please please help on this problem!!!!!!!!!!!!!!!!!!!!!! .pdfFOREVERPRODUCTCHD
SOS Please please please help on this problem!!!!!!!!!!!!!!!!!!!!!! :) Closing Entries and the
Postclosing Trial Balance 192 CHAPTER 6 Mini-Practice Set 1 Service Business Accounting
Cycle Eli\'s Consulting Services ing principles and This project will give you an opportunity to
apply your knowledge of accounting principles and procedures by handling all the accounting
work of Eli\'s Consulting Services for the month o January 2020. Assume that you are the chief
accountant for Eli\'s Consulting Services, During January, the busi ness will use the same types
of records and procedures that you learned about in Chapters 1 through 6. The chart of accounts
for Eli\'s Consulting Services has been expanded to include a few new accounts. Follow the
instructions to complete the accounting records for the month of January. INTRODUCTION
Eli\'s Consulting Services Chart of Accounts Revenue 401 Fees Income Assets 101 Cash 111
Accounts Receivable 121 Supplies 134 Prepaid Insurance 137 Prepaid Rent 141 Equipment 142
Accumulated Depreciation Equipment Liabilities 202 Accounts Payable Expenses 511 Salaries
Expense 514 Utilities Expense 517 Supplies Expense 520 Rent Expense 523 Depreciation
Expense-Equipment 526 Advertising Expense 529 Maintenance Expense 532 Telephone
Expense 535 Insurance Expense Owner\'s Equity 301 Trayton Eli, Capital 302 Trayton Eli,
Drawing 309 Income Summary INSTRUCTIONS 1. Open the general ledger accounts and enter
the balances for January 1, 2020. Obtain the necessary figures from the postclosing trial balance
prepared on December 31, 2019, which appears in Figure 6.3. 2. Analyze each transaction and
record it in the general journal. Use page 3 to begin January\'s transactions. 3. Post the
transactions to the general ledger accounts. 4. Prepare the Trial Balance section of the worksheet.
5. Prepare the Adjustments section of the worksheet. a. Compute and record the adjustment for
supplies used during the month. An inventory taken on January 31 showed supplies of $5,200 on
hand. b. Compute and record the adjustment for expired insurance for the month. c. Record the
adjustment for one month of expired rent of $4,000. d. Record the adjustment for depreciation of
$183 on the old equipment for the month. The first adjustment for depreciation for the new
equipment will be recorded in February, 6. Complete the worksheet. 7. Prepare an income
statement for the month. 8. Prepare a statement of owner\'s equity.
Solution
Journal Entries
Date
Account Title
Debit
Credit
2-Jan
Supplies
7000
Cash
7000
7-Jan
Cash
20000
Accounts receivable
5000
Fees Income
25000
2-Jan
Insurance expense
8400
Cash
8400
12-Jan
Cash
4000
Accounts receivable
4000
Advertising expense
3600
Cash
3600
Cash
20700
Accounts receivable
2300
Fees Income
23000
13-Jan
Cash
4500
Accounts receivable
4500
14-Jan
Cash
750
Supplies
750
20-Jan
Supplies
5000
Accounts Payable
5000
20-Jan
Cash
12500
Accounts receivable
3500
Fees Income
16000
20-Jan
Cash
5600
Accounts receivable
5600
21-Jan
Ma.
Step 12 The task is complete when the quota can be verified. Take a.pdfFOREVERPRODUCTCHD
Step 12 The task is complete when the quota can be verified. Take a Screenshot of file(s), and
submit a print out. Step 13 Set up a static IP address for your eth2 interface on any installation
with following guidelines: The IP address should be 192.168.1.9 The netmask should be
255.255.255.0 The gateway should be set to 192.168.1.1 Step 14 The task is complete when all
interfaces can be pinged successfully. Take a Screenshot of file(s), and submit a print out.
Solution
Linux network interface names start with ethX. First Ethernet network interface name is eth0,
second is eth1,eth2 and so on.
Login as a root user, use su – command:
$ su –
To list or display more information about network interface type command:
# ifconfig | less
To assign an IP address type following command:
# ifconfig eth2 192.168.1.9 up
To take down network interface type following command:
# ifconfig eth2 down
To configure network card using GUI tool type command:
# redhat-config-network
You can also type command setup and select network configuration from menu:
# setup
If you wish to configure network interface eth2 manually then you need to edit files stored in
/etc/sysconfig/network-scripts/ directory. For example here is my sample /etc/sysconfig/network-
scripts/ifcfg-eth2 file for eth2 network interface:
DEVICE=eth2
BOOTPROTO=static
BROADCAST=192.168.1.255
HWADDR=00:0F:EA:91:04:07
IPADDR=192.168.1.9
NETMASK=255.255.255.0
NETWORK=192.168.1.1
ONBOOT=yes
TYPE=Ethernet
Save your changes and exit the text editor.
Restart the network service type following command.
QUESTIONDiscuss how has Web 2.0 changed the behavior of Internet .pdfFOREVERPRODUCTCHD
QUESTION:
Discuss how has Web 2.0 changed the behavior of Internet users? What are the basic tools or
applications that characterize Web 2.0?
While IT provides the platform for this phenomenon, the changing behavior of users represents
the biggest challenge and opportunity for businesses today. Web 2.0 also represents opportunities
for those who understand and master the new way of doing things. Managers who invest the time
to understand and become proficient in new approaches to identifying, communicating and
building relationships with customers online will have a tremendous advantage over managers
who limit themselves to traditional methods.
Blog is short for “Web Log” and is a Web site where users regularly post information for others
to read. Blogs allow readers to comment on each posting. Blogs are a key tool for organizations
that practice content marketing, where valuable information is shared with current or prospective
customers. Bloggers can establish a lot of credibility for themselves and their organizations by
providing helpful information to people who are part of their target market.
Solution
Wikis: Websites that empower clients to contribute, work together and alter webpage content.
Wikipedia is one of the most seasoned and best-known wiki-based locales.
The expanding pervasiveness of Software as a Service (SaaS), web applications and distributed
computing as opposed to privately introduced projects and administrations.
Portable registering, otherwise called nomadicity, the pattern toward clients associating from
wherever they might be. That pattern is empowered by the multiplication of cell phones, tablets
and other cell phones in conjunction with promptly available Wi-Fi systems.
Concoction: Web pages or applications that coordinate reciprocal components from at least two
sources.
Interpersonal interaction: The act of growing the quantity of one\'s business and additionally
social contacts by making associations through people. Interpersonal interaction destinations
incorporate Facebook, Twitter, LinkedIn and Google+.
Communitarian endeavors in view of the capacity to achieve vast quantities of members and
their aggregate assets, for example, crowdsourcing, crowdfunding and crowdsource testing..
Python program with functions that extracts specific characters from.pdfFOREVERPRODUCTCHD
Python program with functions that extracts specific characters from a text file. (e.g. .,#!?)
Solution
data=[] flag=False with open(\'/tmp/myFile.txt\',\'r\') as f: for line in f: if
line.startswith(\'&\'): flag=True if flag: data.append(line) if
line.strip().endswith(\'#!?\'): flag=False print \'\'.join(data).
Please answer the following question and all its parts. Please exp.pdfFOREVERPRODUCTCHD
Please answer the following question and all it\'s parts. Please explain your answer as well!
Thank you in advance for your hardwork! 2) For a 7-way B-tree, answer the following questions
(1 point each) Assignment 4 Heap, Huffman Coding Tree, Multiway Trees Due Date: 9:30am,
Dec. 1 (Hard and E-copy) a) How many keys should each node be able to hold at maximum? b)
What is the minimum number of children an internal node, other than the root, have? c) What is
the maximum number of children can an internal node have?
Solution
In an n-way binary B-tree each node has n children and n-1 keys
each node other than root and leaf nodes have atmost m non empty children and atleast [m/2]
non empty children.
a) 6 keys should each node be able to hold at maximum
b) minimum 4 children can an internal node must have
c) maximum 7 children.
Please help, I cant figure out what I did wrong. Problem 11-2A (Pa.pdfFOREVERPRODUCTCHD
Please help, I can\'t figure out what I did wrong. Problem 11-2A (Part Level Submission) The
stockholders\' equity accounts of Cheyenne Corp. on January 1, 2017, were as follows. Preferred
Stock (7%, $100 par noncumulative, 4,400 shares authorized) Common Stock ($4 stated value,
310,000 shares authorized) Paid-in Capital in Excess of Par Value-Preferred Stock Paid-in
Capital in Excess of Stated Value-Common Stoclk Retained Earnings Treasury Stock (4,400
common shares) $264,000 1,033,33:3 13,200 496,000 683,500 35,200 During 2017, the
corporation had the following transactions and events pertaining to its stockholders equity Feb. 1
Issued 4,620 shares of common stock for $32,340 Mar 20 Purchased 1,250 additional shares of
common treasury stock at $8 per share Oct. 1 Declared a 7% cash dividend on preferred stock,
payable November 1 Nov. 1 Paid the dividend declared on October 1 Dec. Declared a $0.85 per
share cash dividend to common stockholders of record on December 15, payable December 31,
2017 Dec. 31 Determined that net income for the year was $278,200. Paid the dividend declared
on December 1
Solution
I have only corrected the journal entries marked in red
Calculation of cash dividends on Dec1:
No of common stock shares outstanding: (1033,333/4) + 4620 - 4400 - 1250 = 257303.25
<4620 is additional shares issued; 4400 is treasury stock; 1250 is additional treasury stock>
Cash dividend declared = 0.85 per share
Hence, total cash dividend = 257303.25 x 0.85 = $218708
Calculation of cash dividends on Dec 31:
Oct 1 cash dividend (as per journal entry): 18480
Dec 1 cash dividend (as per journal entry): 218708
Total cash dividend: 237188DateAccount titlesDebitCreditFeb
1Cash32340Common stock18480Paid-in Capital in excess of stated value-common
stock13860Dec 1Cash dividends218708Dividends payable218708Dec 31Retained
earnings237188Cash dividends237188Dec 31Dividends payable218708Cash218708.
Know the different types of viruses which have a dsRNA genome A type.pdfFOREVERPRODUCTCHD
Know the different types of viruses which have a dsRNA genome A type of infection caused by
bacteriophages which involves a transition from a lytic to a lysogenic stage. What is the defense
mechanism employed by the T4 phage to prevent excision of its DNA by restriction enzymes?
Know the viruses that use RNA/DNA genomes. Describe in general terms how bacteriophage
lambda regulates the switch between the lytic and lysogenic cycles. Proteins involved? What
factors may influence the transition from a lytic to lysogenic state in lambda phage? Describe in
general terms how bacteriophage lambda regulates the switch between the lytic and lysogenic
cycles. Other Proteins involved? Describe in general terms the strategy used by the ssDNA virus
phiX174 to synthesize its nucleic acids and proteins. Function of the replicative form? Know the
different viruses which have a ssDNA genome Outline the major events involved in plus-strand
RNA viruses and the specific mechanisms used to accomplish each step. Type of template used?
Describe in general terms the strategy used by minus and plus strand RNA viruses to synthesize
their nucleic acids and proteins. Propose how a retrovirus with a single RNA molecule as its
genome might generate multiple proteins from that molecule. Intermediate molecule? What are
cytokines? Roles in immunity? What is the alternate complement pathway? Roles? What are
the functions and locations (blood, tissue etc) of the different immune cells? Mechanisms
employed by the lungs to help protect it from infection What are dendritic cells? Roles they play
in the immune response? Specific vs non-specific Function of lysozyme? Know the different
body areas which act as physical barriers to infection. Know the different physical and
biological defense mechanisms? Examples?
Solution
As per the rule i have answered your first question:
Reoviridae: Reoviridae are currently classified into nine genera. The genomes of these viruses
consist of 10 to 12 segments of dsRNA, each generally encoding one protein. The mature virions
are non-enveloped. Their capsids, formed by multiple proteins, have icosahedral symmetry and
are arranged generally in concentric layers. A distinguishing feature of the dsRNA viruses,
irrespective of their family association, is their ability to carry out transcription of the dsRNA
segments, under appropriate conditions, within the capsid. In all these viruses, the enzymes
required for endogenous transcription are thus part of the virion structure.
Orthoreoviruses: The orthoreoviruses (reoviruses) are the prototypic members of the virus
Reoviridae family and representative of the turreted members, which comprise about half the
genera. Like other members of the family, the reoviruses are non-enveloped and characterized by
concentric capsid shells that encapsidate a segmented dsRNA genome. In particular, reovirus has
eight structural proteins and ten segments of dsRNA. A series of uncoating steps and
conformational changes accom.
In what ways do the experts foresee the use of both virtualization a.pdfFOREVERPRODUCTCHD
In what ways do the experts foresee the use of both virtualization and cloud computing evolving
in the future?
Solution
Use ofVirtualization and cloud computing security :
• Self-defending VM security
• Layered coordinated defenses
• Security optimized for virtual and cloud environments
• Visibility, reporting, and auditing
• Encryption for virtual and cloud environments
• Security that travels with data
Security solutions should offer both agent-less and agent-based security options to provide
flexible deployment alternatives and close security gaps unique to virtualized and cloud
environments.
Encryption addresses a range of security challenges related to virtualization and cloud
computing. Standard 128-, 192-, or 256-bit encryption of storage volumes deters hackers from
prying and thieving, and reduces the risk that the cloud storage devices could be sold or reused
while they still contain confidential or private information. Encryption also greatly reduces the
risk of malicious VM attacks; as long as the encryption key for the data stores have not been
provided, even if rogue VMs reach data stores, volumes are unmountable and unreadable.
Encryption with enterprise-controlled key management enables IT to comply with security best
practices, internal governance, and external regulation. Data is kept secure, and the key
management solution can provide monitoring, reporting, and auditing capabilities that provide
visibility into data access. As a result, enterprises realize a significant reduction in the scope of
compliance audits.
Encryption for virtual and cloud environments:
In summary, best practices for encryption for virtual and cloud data protection include: •
Integration with leading cloud service providers and virtual environments. • Policy-based key
management that determines where and when encrypted data can be accessed. • Identity-based
and integrity-based server validation to determine which servers can access secure storage
volumes and whether security is up-to-date on those servers prior to data access. • Business
control of encryption keys, either on-site or through a separate SaaS service, to maintain a strict
separation of duties between the business and cloud service provider..
In the Meselson Stahl experiment, E. coli was grown for many generati.pdfFOREVERPRODUCTCHD
In the Meselson Stahl experiment, E. coli was grown for many generations in a medium
\"heavy\" nitrogen^15 N. Once the bacterial DNA was uniformly labeled, the bacteria were
transferred to a medium containing the \"light\" nitrogen, ^14N. After ONE round of replication,
DNA was extracted and subjected to density gradient centrifugation in cesium chloride. Upon
inspection, there was only a single band of intermediate density DNA in the gradient. This
proved that DNA replication CANNOT be conservative dispersive semiconservative
conservative of dispersive continuous.
Solution
Answer is Dispersive replication.
Dispersive replication would result in double-stranded DNA with both strands having mixtures
of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density..
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
implement the ListLinked ADT (the declaration is given in ListLinked.pdf
1. implement the ListLinked ADT (the declaration is given in ListLinked.h)(60 points)
- implement the following operations:
- constructor, assignment operator, destructor
- insert, remove, replace, clear
- isFull, isEmpty
- gotoBeginning, gotoEnd, gotoNext, gotoPrior, getCursor
implement the function moveToBeginning() that removes the data item marked by the cursor and
reinserts it at the beginning of the list
- implement the function insertBefore(..) that will insert the new data item before the cursor or if
the list is empty as the first element of the list
#include "ListLinked.h"
// ListNode member functions
template
List::ListNode::ListNode(const DataType& nodeData, ListNode* nextPtr)
{
this->dataItem = nodeData;
this->next = nextPtr;
}
// List member functions
template
List::List(int ignored = 0)
{
}
template
List::List(const List& other)
{
}
template
List& List::operator=(const List& other)
{
}
template
List::~List()
{
}