operating system linux,ubuntu,Mac#include iostream #include .pdf

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//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"<.

Education

$//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) {$

$clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << " "; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<"Please Enter 3 argument"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num)$

${ if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << " "; head = head->nextNode; }$

std::cout << endl;
}
int main(int argc, char *argv[])
{
struct Node* head = NULL,*temp;
srand(time(0));
if ( argc != 3 ) // argc should be 2 for correct execution
// We print error message
std::cout<<"Please Enter 3 argument"<

Similar to operating system linux,ubuntu,Mac#include iostream #include .pdf

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.

take the following code and give details of what each line of code i.pdf

fastechsrv

// main public class AssignmentThree { public static void main(String[] args) { List myList = new List(15); // Cause List Empty Message myList.removeFront(); myList.removeRear(); myList.removeItem(\"a\"); // Cause Not found message myList.addToFront(\"x\"); myList.removeItem(\"y\"); myList.removeItem(\"x\"); myList.addAfterItem(\"x\", \"z\"); myList.addBeforeItem(\"x\", \"z\"); // Normal behavior myList.addToFront(\"not.\"); myList.addToFront(\"or\"); myList.addToRear(\"is\"); myList.addToRear(\"try.\"); myList.addAfterItem(\"is\", \"no\"); myList.addBeforeItem(\"is\", \"There\"); myList.addToFront(\"Do\"); myList.addAfterItem(\"or\", \"do\"); myList.print(\"Original list\"); myList.printSorted(\"Sorted Original List\"); sop(\"\ Front is \" + myList.getFront()); sop(\"Rear is \" + myList.getRear()); sop(\"Count is \" + myList.askCount()); sop(\"Is There present? \" + myList.isPresent(\"There\")); sop(\"Is Dog present? \" + myList.isPresent(\"Dog\")); myList.addToFront(\"junk\"); myList.addToRear(\"morejunk\"); myList.addAfterItem(\"or\", \"moremorejunk\"); myList.print(\"List with junk\"); sop(\"Count is \" + myList.askCount()); myList.removeFront(); myList.removeRear(); myList.removeItem(\"moremorejunk\"); myList.print(\"List with junk removed\"); sop(\"Count is \" + myList.askCount()); sop(\"\"); // Cause List Full message for(int ii = 0; ii < 10; ++ii) { myList.addToFront(DUMMY); } myList.addToRear(DUMMY); myList.addBeforeItem(\"no\", DUMMY); myList.addAfterItem(\"There\", DUMMY); myList.print(\"After filling List\"); sop(\"Count is \" + myList.askCount()); while(myList.isPresent(DUMMY)) myList.removeItem(DUMMY); myList.print(\"After removing \" + DUMMY ); sop(\"Count is \" + myList.askCount()); } private static void sop(String s) { System.out.println(s); } private static final String DUMMY = \"dummy\"; } Solution Following is MyList implementation for your output as: Java Code: import java.util.Comparator; public class MyList { private Node start; private Node end; private int size; private int capacity; public MyList(int capacity) { start = null; end = null; size = 0; this.capacity = capacity; } /* Function to check if list is empty */ public boolean isEmpty() { return start == null; } public void removeFront() { if (isEmpty()) { System.out.println(\"List Empty\"); return; } else { start = start.next; size--; return; } } public void removeRear() { if (size == 0) { System.out.println(\"List Empty\"); return; } if (start.next == null) { start=null; end=null; size=0; } Node node = start.next; while (node.next != null) { node = node.next; } node.next = null; end = node; size--; } public void removeItem(String key) { if (size == 0) { System.out.println(\"List Empty\"); return; } if (start.next == null) { if (start.key.equals(key)) { start=null; end=null; size=0; } else { System.out.println(\"\"); } return; } Node node = start; Node nodeNext = node.next; while (nodeNext.next != null) { if (nodeNext.key.equals(key)) { node.next=nodeNext.next; }.

mainpublic class AssignmentThree { public static void ma.pdf

fathimafancyjeweller

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.

Write a program that accepts an arithmetic expression of unsigned in.pdf

JUSTSTYLISH3B2MOHALI

reverse the linked list (2,4,8,10) by: stack; iteration; recursion. Using C++ Solution #include<iostream> #include<stdio.h> #include<stdlib.h> #includ<stack> using namespace std; struct Node { int value; struct Node* next; }node; struct Node* reverse(struct Node* head) {Â Â // Iteration Method struct Node* current,*prev; current = head; prev = NULL; while(current!=NULL) { next = current->next; current->next = prev; prev = current; current = next; } head = prev;- return head; } struct Node* insert(Node* head,int value) { Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->value = value; temp->next = NULL; if(head = = NULL) head = temp; else { Node*temp1=head; while(temp1->next!=NULL) temp1 = temp1->next; temp1->next = temp; } return head; } public void print(Node* head) { while(head!=NULL) { cout<<head->value; head = head->next; } } struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method if(head==NULL || head->next ==NULL) return; Node*first = head; Node *rest = head->next; if(rest !=NULL) return; recursiveReverse(rest); first ->next->next = first; first->next =NULL; head = rest; return head; } void Reverse(struct Node* head)Â Â // stack Implementation { stack<node*> Stack; node *traverse = head; while(traverse!=NULL) { Stack.push(traverse); traverse = traverse->next;Â Â } node *temp = Stack.top(); head = temp; Stack.pop(); while(!Stack.empty()) { temp->next = Stack.top(); Stack.pop(); temp = temp->next;Â Â } temp->next = NULL; return Stack; } int main() { struct Node* head =NULL; head = insert(head,2); head = insert(head,4); head = insert(head,8); head = insert(head,10); print(head); head= reverse(head); print(head); head = recursiveReverse(head); print(head); head = void Reverse(head); void print(head); return 0; } .

reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx

acarolyn

#include <stdio.h> #include <malloc.h> #define ISEMPTY printf(\"\ EMPTY LIST:\"); /* * Node Declaration */ struct node ; snode* create_node(int); void insert_node_first(); void insert_node_last(); void insert_node_pos(); void sorted_ascend(); void delete_pos(); void search(); void update_val(); void display(); void rev_display(snode *); typedef struct node snode; snode *newnode, *ptr, *prev, *temp; snode *first = NULL, *last = NULL; /* * Main :contains menu */ int main() whereas (ans == \'Y\'||ans == \'y\') value = val; newnode->next = NULL; {return|come back|come\" id=\"tip_27\">come newnode; } } /* * Inserting Node initially */ void insert_node_first() worker = first; initial = newnode; first->next = temp; } printf(\"\ ----INSERTED----\"); } /* * Inserting Node eventually */ void insert_node_last() printf(\"\ ----INSERTED----\"); } /* * Inserting Node at position */ void insert_node_pos() if (pos == 1) { if (first == last && initial == NULL) { initial = last = newnode; first->next = NULL; last->next = NULL; } else worker = first; initial = newnode; first->next = temp; } printf(\"\ Inserted\"); } else if (pos>1 && pos<=cnt) prev->next = newnode; newnode->next = ptr; printf(\"\ ----INSERTED----\"); } else } } printf(\"\ ---Sorted List---\"); for (ptr = first;ptr != NULL;ptr = ptr->next) if ( Solution #include <stdio.h> #include <malloc.h> #define ISEMPTY printf(\"\ EMPTY LIST:\"); /* * Node Declaration */ struct node ; snode* create_node(int); void insert_node_first(); void insert_node_last(); void insert_node_pos(); void sorted_ascend(); void delete_pos(); void search(); void update_val(); void display(); void rev_display(snode *); typedef struct node snode; snode *newnode, *ptr, *prev, *temp; snode *first = NULL, *last = NULL; /* * Main :contains menu */ int main() whereas (ans == \'Y\'||ans == \'y\') value = val; newnode->next = NULL; {return|come back|come\" id=\"tip_27\">come newnode; } } /* * Inserting Node initially */ void insert_node_first() worker = first; initial = newnode; first->next = temp; } printf(\"\ ----INSERTED----\"); } /* * Inserting Node eventually */ void insert_node_last() printf(\"\ ----INSERTED----\"); } /* * Inserting Node at position */ void insert_node_pos() if (pos == 1) { if (first == last && initial == NULL) { initial = last = newnode; first->next = NULL; last->next = NULL; } else worker = first; initial = newnode; first->next = temp; } printf(\"\ Inserted\"); } else if (pos>1 && pos<=cnt) prev->next = newnode; newnode->next = ptr; printf(\"\ ----INSERTED----\"); } else } } printf(\"\ ---Sorted List---\"); for (ptr = first;ptr != NULL;ptr = ptr->next) if (.

#include <stdio.h> #include <malloc.h> #define ISEMP.pdf

harihelectronicspune

DS Code (CWH).docx

KamalSaini561034

//tested on eclipse /*************DoublyLinkedList class*******************/ public class DoublyLinkedList { /*Data member variable declaration*/ private Node head; private int count; /**default constructor*/ public DoublyLinkedList() { /*Initializing variable*/ head=null; count =0; } /*Insert at beginning in doubly linked list*/ public void insertBeginning(String name){ /*checking head is null then we will add node at head*/ if(head==null){ head=new Node(name, null, null); count++; }else{ /*else create temp node and setting next to head * and then making temp to head*/ Node temp=new Node(name,null,null); temp.setNext(head); head.setPrev(temp); head=temp; count++; } } public void printList(){ Node temp=head; /*Printing element of doubly list*/ while(temp!=null){ System.out.println(temp.getData()); temp=temp.getNext(); } } /*Returning count from doubly linked list*/ public int count(){ return count; } /*deleting specific node from doubly linked list*/ public void delete(String name){ Node current=head,previous=head; /*if it is head node to be deleted*/ if(head.getData().equals(name)){ current=head.getNext(); current.setPrev(null); head=current; previous=head; }else{ /*If node in middle of list*/ while(current.getNext()!=null){ if(current.getData().equals(name)){ current.getNext().setPrev(previous); previous.setNext(current.getNext()); count--; break; } previous=current; current=current.getNext(); } /*If node is at last*/ if(current.getNext()==null){ if(current.getData().equals(name)){ previous.setNext(null); count--; } } } } /*deleting all nodes from list by assigning head to null * and memory releasing take care by garbage collector*/ public void clear(){ head=null; count=0; } /*Inserting element in alphabetic order*/ public void insertAlpha(String name){ Node current=head,previous=head; /*creating new node*/ Node newNode=new Node(name,null,null); /*if head node is null*/ if(head==null){ head=newNode; count++; }else{ /*please read about compareTo() method about returning value*/ if(head.getData().compareTo(name)>0){ newNode.setNext(head); head=newNode; count++; }else{ /*condition for getting position of node to be inserted*/ while(current!=null&&current.getData().compareTo(name)<=0){ previous=current; current=current.getNext(); } /*if node need to be inserted at last*/ if(current==null){ previous=newNode; }else{ /*If node need to inserted in middle*/ newNode.setNext(current); previous.setNext(newNode); current.setPrev(newNode); newNode.setPrev(previous); } count++; } } } /*Main method start*/ public static void main(String args[]){ /*creating object of doubly linked list*/ DoublyLinkedList doublyLinkedList=new DoublyLinkedList(); doublyLinkedList.insertBeginning(\"sunita\"); doublyLinkedList.insertBeginning(\"lalchand\"); /*calling alphabetic order method*/ doublyLinkedList.insertAlpha(\"ramsingh\"); doublyLinkedList.insertBeginning(\"Lokesh\"); doublyLinkedList.insertBeginning(\"Pritam\"); /*****printing list********/ doublyLinkedList.printL.

tested on eclipseDoublyLinkedList class.pdf

shanki7

write recursive function that calculates and returns the length of a linked list Solution // C code to determine length of linked list recursively #include #include // node of linked list struct node { int key; struct node* next; }; // push key to linked list void push(struct node** root, int new_element) { struct node* nodeNew = (struct node*) malloc(sizeof(struct node)); nodeNew->key = new_element; nodeNew->next = (*root); (*root) = nodeNew; } // recursively get length of linked list int lengthLinkedlist(struct node* root) { // base case if (root == NULL) return 0; // recursively call the function return 1 + lengthLinkedlist(root->next); } int main() { struct node* root = NULL; push(&root, 11); push(&root, 34); push(&root, 4); push(&root, 23); push(&root, 14); push(&root, 88); push(&root, 56); push(&root, 84); push(&root, 6); /* Check the count function */ printf(\"Length of Linked list: %d\ \", lengthLinkedlist(root)); return 0; } /* output: Length of Linked list: 9 */.

write recursive function that calculates and returns the length of a.pdf

arpitcomputronics

/* Task 2: Debugging a program with stacks, queues, and doubly-linked lists There are a number of errors in the following program. All errors are located in main() and structure definitions. Function declarations and definitions are correct! Locate all errors, fix them (as shown below), run the program and save its output as a comment at the end of the source file. Example: int num = 10; int *ptr; num = &ptr; // <== Error: Comment the line and write the correct line below // Write a short justification where appropriate // num = &ptr; // Error #1 ptr = # Name: */ #include #include #include #include #define DUMMY_TRAILER '\177' // octal ASCII code of the // last character in the ASCII table #define NUM_CITIES 10 typedef struct { char name[12]; int temperature[5]; } CITY; // Stack and Queue Node typedef struct node NODE; struct node { CITY city; node *next; }; // Doubly Linked List Node typedef struct d_node D_NODE; struct d_node { CITY city; NODE *forw; NODE *back; }; // Stack Functions NODE *push(NODE *stack, const CITY *pStu); NODE *pop(NODE **stack); // Queue Functions void enqueue(NODE **queue, NODE **rear, const CITY *pStu); NODE *dequeue(NODE **queue, NODE **rear); // Doubly Linked List Functions D_NODE *init_list(void); int insert(D_NODE *list, const CITY *pStu); void traverse_forw(D_NODE *list); void traverse_back(D_NODE *list); // Other Functions void printCity(const CITY *pCity); int main (void) { CITY cList[NUM_CITIES] = { {"Cupertino", {88, 89, 87, 85, 89}}, {"Flagstaff", {81, 80, 88, 89, 89}}, {"Los Angeles", {87, 88, 89, 89, 90}}, {"Philadelphia", {96, 99, 99, 90, 95}}, {"Phoenix", {106, 109, 109, 100, 105}}, {"Portland", {89, 90, 85, 89, 90}}, {"Reno", {108, 105, 109, 100, 108}}, {"Salem", {85, 90, 85, 89, 90}}, {"Tucson", {107, 100, 109, 100, 108}}, {"Yreka", {101, 109, 100, 108, 109}} }; NODE *stack = NULL; NODE *top = NULL; NODE *queue = NULL, *rear = NULL; NODE *front; D_NODE *list; list = init_list(); // build stack and queue with data from an array of CITY structures srand((unsigned int)time(NULL)); int count = rand() % 10; for ( int n = 0; n < count; n++) { int i = rand() % NUM_CITIES; int duplicate = insert(list, &cList[i]); if(duplicate) { // already in the list! push(stack, &cList[i]); enqueue(&queue, &rear, cList[i]); } } // display list printf("\nLIST contents (forwards):\n"); traverse_forw(list); printf("\nLIST contents (backwards):\n"); traverse_back(list); // display stack if (top) { printf("\nSTACK contents from top to bottom:\n"); while ((top = pop(stack))) { printCity(top->city); } } else printf ("Empty Stack!\n"); // display queue if (front) { printf("\nQUEUE contents from front to rear:\n"); while ((front = dequeue( queue, rear))) { printCity(front->city); } } else printf ("Empty Queue!\n"); return 0; } /*************************************************** Displays the fileds of a CIS_CLASS structure Pre pCls - a pointer to a CIS_CLASS structure Post */ void printCity(const CITY *pCity) { printf("%-20s %3d\n",.

-- Task 2- Debugging a program with stacks- queues- and doubly-linked.docx

Adamq0DJonese

Linked lists

George Scott IV

#include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2); p1->insertFront(\'a\'); p1->print(); system(\"pause\"); return 0; } Solution #include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2.

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ankitmobileshop235

coding in C; Create a function called reverseList that takes the head of a linked list, reverses the order of all the nodes. For example, if the list contained 1, 2, 3, 4 in its nodes, the list will now contain 4, 3, 2, 1. please add a main test. Solution Please find the required program below: #include<stdio.h> #include<stdlib.h> struct node { int data; struct node* next; }; void push(struct node** head, int new_element) { struct node* new_node = (struct node*) malloc(sizeof(struct node)); new_node->data = new_element; new_node->next = (*head); (*head) = new_node; } void printList(struct node *head) { struct node *temp = head; while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } void reverseList(struct node** head) { struct node* prev = NULL; struct node* current = *head; struct node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head = prev; } int main() { struct node* head = NULL; push(&head, 12); push(&head, 14); push(&head, 5); push(&head, 8); printf(\"\ Linked list Before Reversing :\ \"); printList(head); reverseList(&head); printf(\"\ Linked list After Reversing :\ \"); printList(head); } -------------------------------------------------------------------------------- OUTPUT: Linked list Before Reversing : 8Â Â 5Â Â 14Â Â 12 Linked list After Reversing : 12Â Â 14Â Â 5Â Â 8 .

coding in C- Create a function called reverseList that takes the head.docx

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Program In C You are required to write an interactive C program that prompts the user for commands, accepts commands from the keyboard (stdin) and executes those commands. When a command requires output, it must be written to stdout. The program must continue to accept and process commands until the user types the end command ourput, it must be writen to stdout. The program must continue to accept and proces.commands The program deals with linked lists. Each node of such a list contains a string of length at most 10, a positive integer (i.e., an integer value 2 1) and a pointer to the next node of the list. For any node, the string and the integer stored in that node will be referred to as the symbol and count for that node respectively. At all times, the list must satisfy the following two important properties. 1. The symbols appearing in the list are all distinct; that is, no two nodes have the same symbol. 2. When the list is scanned from left to right, the counts must be in non-increasing order. An example of such a linked list is shown below. Search 4 Maxval4 begin/ Loop head Solution /**********************************************/ /****** NAMES ******************** COMMANDS ***/ /* Insert Command : ins str */ /* Delete Command : del str */ /* Forced Delete Command : fde val */ /* Print Statistics Command : pst */ /* Print List Command : prl */ /* Print using Count Range Command: pcr v1 v2 */ /* Print Prefix Command : ppr str */ /* Print Suffix Command : psu str */ /* End Command : end */ /***********************************************/ #include #include #include #define CSIZE 4 #define SSIZE 11 /* Struct for each node in the linked list */ struct node { int count; char symbol[SSIZE]; struct node *next; }; struct node *head; /* Prototypes for the functions */ void insert(char *); void delete(char *); void forced_delete(int value); void print_stats(); void print_list(); void print_count(int, int); void print_prefix(char pre[]); void print_suffix(char suf[]); void sort_list(); /**************** ** Main Method ** ****************/ int main(void) { int count; char symbol[SSIZE]; char cmd[CSIZE]; int v1; int v2; /* Enter commands */ printf(\"Command? \"); fflush(stdout); scanf(\"%s\", cmd); /* Check for command*/ while(strcmp(cmd,\"end\") != 0) { /* check which function to use */ if(strcmp(cmd,\"ins\") == 0) { scanf(\"%s\", symbol); insert(symbol); } if(strcmp(cmd,\"del\") == 0) { scanf(\"%s\", symbol); delete(symbol); } else if(strcmp(cmd,\"fde\")==0){scanf(\"%d\",&count); forced_delete(count); } if(strcmp(cmd,\"pst\") == 0) { print_stats();} if(strcmp(cmd,\"prl\") == 0) { print_list(); } if(strcmp(cmd,\"pcr\") == 0) { scanf(\"%d %d\",&v1,&v2); print_count(v1,v2); } if(strcmp(cmd,\"ppr\") == 0) { scanf(\"%s\", symbol); print_prefix(symbol);} if(strcmp(cmd,\"psu\") == 0) { scanf(\"%s\", symbol); print_suffix(symbol);} printf(\"Command? \"); fflush(stdout); scanf(\"%s\", cmd); }// End of While Loop return 0; }// End of main method /*********.

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Implement of c & its coding programming by sarmad baloch

Sarmad Baloch

How do you stop infinite loop? Because I believe that it is making an infinite circular list. c++ code: Here is the list class: #ifndef LIN_J_LIST #define LIN_J_LIST typedef unsigned int uint; #include #include using namespace std; /** * a simplified generic singly linked list class to illustrate C++ concepts * @author Jerry Lin * @version 2/17/17 */ template< typename Object > class List { private: /** * A class to store data and provide a link to the next node in the list */ class Node { public: /** * The constructor * @param data the data to be stored in this node */ explicit Node( const Object & data ) : data{ data }, next{ nullptr } {} Object data; Node * next; }; public: /** * The constructor for an empty list */ List() : size{ 0 }, first{ nullptr }, last{ nullptr } {} /** * the copy constructor-creates and copy the list */ List( List && rhs ) = delete; List( const List & rhs ) { count = 0; size = 0; if(rhs.size != 0) { push_front(rhs.front()); Node * current = rhs.first; Node * tempNode = first; size++; while(current->next != nullptr) { current = current->next; Node *newNode = new Node(current->data); //transfer the data. basic op. count++; tempNode->next = newNode; last = newNode; tempNode = tempNode->next; size++; } } // you document and implement this method } /** * the operator= method-copies the list */ List & operator=( List && rhs) = delete; List & operator=( const List & rhs ) { count = 0; size = 0; if( size != 0 ) { Node * current = first; Node * temp; while( current != nullptr ) { temp = current; current = current->next; delete temp; } } if(rhs.size!= 0) { push_front(rhs.front()); Node * current = rhs.first; Node * tempNode = first; //create a temporary to store size++; while(current -> next != nullptr) { current = current->next; Node *newNode = new Node(current->data); //transfer the data. basic op count++; tempNode->next = newNode; last = newNode; tempNode = tempNode -> next; size++; } } return *this; // you document and implement this method } /** * accessor * @return count */ int get_count() const { return count; } /** * The destructor that gets rid of everything that\'s in the list and * resets it to empty. If the list is already empty, do nothing. */ ~List() { if( size != 0 ) { Node * current = first; Node * temp; while( current != nullptr ) { temp = current; current = current->next; delete temp; } } } /** * Put a new element onto the beginning of the list * @param item the data the new element will contain */ void push_front( const Object& item ) { Node * new_node = new Node( item );//basic op. if(is_empty()) { last = new_node; } else { new_node->next = first; } first = new_node; size++; /* you complete the rest */ } /** * Remove the element that\'s at the front of the list. Causes an * assertion error if the list is empty. */ void pop_front() { assert( !is_empty() ); Node * temp = first; if( first == last ) { first = last = nullptr; } else { first = first->next; } delete temp; size--; } /** * Accessor to return the da.

How do you stop infinite loop Because I believe that it is making a.pdf

feelinggift

* Program to insert in a sorted list */ #include #include /* Link list node */ struct node { int data; struct node* next; }; /* function to insert a new_node in a list. Note that this function expects a pointer to head_ref as this can modify the head of the input linked list (similar to push())*/ void sortedInsert(struct node** head_ref, struct node* new_node) { struct node* current; /* Special case for the head end */ if (*head_ref == NULL || (*head_ref)->data >= new_node->data) { new_node->next = *head_ref; *head_ref = new_node; } else { /* Locate the node before the point of insertion */ current = *head_ref; while (current->next!=NULL && current->next->data < new_node->data) { current = current->next; } new_node->next = current->next; current->next = new_node; } } /* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */ /* A utility function to create a new node */ struct node *newNode(int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; new_node->next = NULL; return new_node; } /* Function to print linked list */ void printList(struct node *head) { struct node *temp = head; while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } /* Drier program to test count function*/ int main() { /* Start with the empty list */ struct node* head = NULL; struct node *new_node = newNode(5); sortedInsert(&head, new_node); new_node = newNode(10); sortedInsert(&head, new_node); new_node = newNode(7); sortedInsert(&head, new_node); new_node = newNode(3); sortedInsert(&head, new_node); new_node = newNode(1); sortedInsert(&head, new_node); new_node = newNode(9); sortedInsert(&head, new_node); printf(\"\ Created Linked List\ \"); printList(head); return 0; } public class Listnode { //*** fields *** private Object data; private Listnode next; //*** methods *** // 2 constructors public Listnode(Object d) { this(d, null); } public Listnode(Object d, Listnode n) { data = d; next = n; } // access to fields public Object getData() { return data; } public Listnode getNext() { return next; } // modify fields public void setData(Object ob) { data = ob; } public void setNext(Listnode n) { next = n; } } Thsi below program print out all the varibles in SLL #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Takes head pointer of the linked list and index as arguments and return data at index*/ int GetNth(struct node* head, int index) { struct node* current = head; int .

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1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of humar resource during the period of industrialization era or machine age worlwide. It is a mix of social drama and comedy. The content is presented in a satire. How did people survived the great depression of the world is being documented here. The conditions werre miserable like acute shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic representation of what happens to a human being caught up in the mechanical bureaucracy of the intellectual state. 2) Economically it could be understood that what miser situations could depression bring up. Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for survival. At the time of the great depression Americans had to fight for what little they had, and as the movie showed many people went hungry. Modern Times looks toward the future, but not with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme: the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy, urbanization, and law enforcement. 3) From the movie a prospect employee should not forget about the social ills and failings of own day even if present age is seems far better than that was shown in the film. A dilbert-like early scene shows the Tramp trying to take a break during work hours in the washroom — only to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order him to quit stalling and get back to work. Modern audiences watching this scene may reflect ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously there’s nothing new about widespread poverty and want, but unemployment as we know it today is a comparatively recent development — a byproduct of the modern workforce economy that requires everyone to \"get a job.\" 4) The major ethical dilemma consists whether out of povertty stealing is right or wrong. However this movie glorifies the struggle of the poor rather than vilifying them for stealing food to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor steal to eat it is not criminal, just survival. Minor ones include Solution 1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of humar resource during the period of industrialization era or machine age worlwide. It is a mix of social drama and comedy. The content is presented in a satire. How did people survived the great depression of the world is being documented here. The conditions werre miserable like acute shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic representation of what happens to a human being caught up in the mechanical bureaucracy of the intellectual state. 2) Economically it could be understood that what miser situations could depress.

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