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  • Java: Is there a way to efficiently insert or remove many elements from the middle of a LinkedList?

    - by allyourcode
    I was expecting to find this in Java's LinkedList, since the point of linked lists is to be able to efficiently insert (and remove) anywhere (assuming you have some kind of pointer to the location where you want to insert or remove). I'm not finding anything in the API though. Am I overlooking something? The closest thing I can find to this are the add and remove method in ListIterator. This has some limitations though. In particular, other iterators become invalid as soon as the underlying LinkedList is modified via remove, according to the API. This is born out in my tests as well; the following program results in a IllegalStateException: import java.util.*; public class RemoveFromLinkedList { public static void main(String[] args) { LinkedList<Integer> myList= new LinkedList<Integer>(); for (int i = 0; i < 10; ++i) { myList.add(i); } ListIterator<Integer> i1 = myList.listIterator(); ListIterator<Integer> i2 = myList.listIterator(); for (int i = 0; i < 3; ++i) { i1.next(); i2.next(); } System.out.println("i1.next() should be 3: " + i1.next()); i1.remove(); i1.remove(); // Exception! System.out.println("i2.next() should be 5: " + i2.next()); } } Ideally, what I'm expecting is something like this: // In my imagination only. This is the way Java actually works, afaict. // Construct two insertion/deletion points in LinkedList myLinkedList. myIterator = myLinkedList.iterator(); for (...) { myIterator.next(); } start = myIterator.clone(); for (...) { myIterator.next(); } // Later... after = myLinkedList.spliceAfter(myIterator, someOtherLinkedList); // start, myIterator, and after are still all valid; thus, I can do this: // Removes everything I just spliced in, as well as some other stuff before that. myLinkedList.remove(start, after); // Now, myIterator is invalid, but not start, nor after. C++ has something like this for its list class (template). Only iterators pointing to moved elements become invalidated, not ALL iterators.

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  • Where is the mistake ?

    - by mr.bio
    Hi ... i am implementing a simple linked list in c++. I have a mistake and i don't see it :( #include <stdexcept> #include <iostream> struct Node { Node(Node *next, int value): next(next), value(value) { } Node *next; int value; }; class List { Node *first; // Erstes Element , 0 falls die Liste leer ist int len; // Laenge der liste Node *nthNode(int index); // Hilfsfunktion : O( index ) public: // Default - Konstruktor ( Laenge 0): O (1) List():first(0),len(0){ } // Copy - Konstruktor : O(other.len) List(const List & other){ }; // Zuweisungs - Operator O(len +other.len) List &operator=(const List &other) { clear(); if(!other.len) return *this; Node *it = first = new Node(0,other.first->value); for (Node *n = other.first->next; n; n = n->next) { it = it->next = new Node(0, n->value); } len = other.len; return *this; } // Destruktor ( gibt den Speicher von allen Nodes frei ): O( len ) ~List(){ }; // Haengt der Liste ein Element hinten an: O( len ) void push_back(int value){ }; // Fuegt am Anfang der Liste ein Element ein : O (1) void push_front(int value){ Node* front = new Node(0,value); if(first){ first = front; front->next = 0; }else{ front->next = first; first = front; } len++; }; // gibt eine Referenz auf das index -te Element zurueck : O( index ) int &at(int index){ int count = 0 ; int ret ; Node *it = first; for (Node *n = first->next; n; n = n->next) { if(count==index) ret = n->value; count++; } return ret ; }; // Entfernt alle Elemente : O(len) void clear(){ }; // Zeigt alle Elemente an: hier : O( len * len ) void show() { std::cout << " List [" << len << " ]:{ "; for (int i = 0; i < len; ++i) { std::cout << at(i) << (i == len - 1 ? '}' : ','); } std::cout << std::endl; } }; /* * */ int main() { List l; // l. push_back(1); // l. push_back(2); l. push_front(7); l. push_front(8); l. push_front(9); l.show(); // List(l). show(); } it works ... but the output is : List [3 ]:{ 0,134520896,9484585}

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  • What's the right way to do mutable data structures (e.g., skip lists, splay trees) in F#?

    - by dan
    What's a good way to implement mutable data structures in F#? The reason I’m asking is because I want to go back and implement the data structures I learned about in the algorithms class I took this semester (skip lists, splay trees, fusion trees, y-fast tries, van Emde Boas trees, etc.), which was a pure theory course with no coding whatsoever, and I figure I might as well try to learn F# while I’m doing it. I know that I “should” use finger trees to get splay tree functionality in a functional language, and that I should do something with laziness to get skip-list functionality, etc. , but I want to get the basics nailed down before I try playing with purely functional implementations. There are lots of examples of how to do functional data structures in F#, but there isn’t much on how to do mutable data structures, so I started by fixing up the doubly linked list here into something that allows inserts and deletes anywhere. My plan is to turn this into a skip list, and then use a similar structure (discriminated union of a record) for the tree structures I want to implement. Before I start on something more substantial, is there a better way to do mutable structures like this in F#? Should I just use records and not bother with the discriminated union? Should I use a class instead? Is this question "not even wrong"? Should I be doing the mutable structures in C#, and not dip into F# until I want to compare them to their purely functional counterparts? And, if a DU of records is what I want, could I have written the code below better or more idiomatically? It seems like there's a lot of redundancy here, but I'm not sure how to get rid of it. module DoublyLinkedList = type 'a ll = | None | Node of 'a ll_node and 'a ll_node = { mutable Prev: 'a ll; Element : 'a ; mutable Next: 'a ll; } let insert x l = match l with | None -> Node({ Prev=None; Element=x; Next=None }) | Node(node) -> match node.Prev with | None -> let new_node = { Prev=None; Element=x; Next=Node(node)} node.Prev <- Node(new_node) Node(new_node) | Node(prev_node) -> let new_node = { Prev=node.Prev; Element=x; Next=Node(node)} node.Prev <- Node(new_node) prev_node.Next <- Node(new_node) Node(prev_node) let rec nth n l = match n, l with | _,None -> None | _,Node(node) when n > 0 -> nth (n-1) node.Next | _,Node(node) when n < 0 -> nth (n+1) node.Prev | _,Node(node) -> Node(node) //hopefully only when n = 0 :-) let rec printLinkedList head = match head with | None -> () | Node(x) -> let prev = match x.Prev with | None -> "-" | Node(y) -> y.Element.ToString() let cur = x.Element.ToString() let next = match x.Next with | None -> "-" | Node(y) -> y.Element.ToString() printfn "%s, <- %s -> %s" prev cur next printLinkedList x.Next

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  • why it throws index out of bounds exception??

    - by Johanna
    Hi I want to use merge sort for sorting my doubly linked list.I have created 3 classes(Node,DoublyLinkedList,MergeSort) but it will throw this exception for these lines: 1.in the getNodes method of DoublyLinkedList---> throw new IndexOutOfBoundsException(); 2.in the add method of DoublyLinkedList-----> Node cursor = getNodes(index); 3.in the sort method of MergeSort class------> listTwo.add(x,localDoublyLinkedList.getValue(x)); 4.in the main method of DoublyLinkedList----->merge.sort(); this is my Merge class:(I put the whole code for this class for beter understanding) public class MergeSort { private DoublyLinkedList localDoublyLinkedList; public MergeSort(DoublyLinkedList list) { localDoublyLinkedList = list; } public void sort() { if (localDoublyLinkedList.size() <= 1) { return; } DoublyLinkedList listOne = new DoublyLinkedList(); DoublyLinkedList listTwo = new DoublyLinkedList(); for (int x = 0; x < (localDoublyLinkedList.size() / 2); x++) { listOne.add(x, localDoublyLinkedList.getValue(x)); } for (int x = (localDoublyLinkedList.size() / 2) + 1; x < localDoublyLinkedList.size(); x++) { listTwo.add(x, localDoublyLinkedList.getValue(x)); } //Split the DoublyLinkedList again MergeSort sort1 = new MergeSort(listOne); MergeSort sort2 = new MergeSort(listTwo); sort1.sort(); sort2.sort(); merge(listOne, listTwo); } public void merge(DoublyLinkedList a, DoublyLinkedList b) { int x = 0; int y = 0; int z = 0; while (x < a.size() && y < b.size()) { if (a.getValue(x) < b.getValue(y)) { localDoublyLinkedList.add(z, a.getValue(x)); x++; } else { localDoublyLinkedList.add(z, b.getValue(y)); y++; } z++; } //copy remaining elements to the tail of a[]; for (int i = x; i < a.size(); i++) { localDoublyLinkedList.add(z, a.getValue(i)); z++; } for (int i = y; i < b.size(); i++) { localDoublyLinkedList.add(z, b.getValue(i)); z++; } } } and just a part of my DoublyLinkedList: private Node getNodes(int index) throws IndexOutOfBoundsException { if (index < 0 || index > length) { throw new IndexOutOfBoundsException(); } else { Node cursor = head; for (int i = 0; i < index; i++) { cursor = cursor.getNext(); } return cursor; } } public void add(int index, int value) throws IndexOutOfBoundsException { Node cursor = getNodes(index); Node temp = new Node(value); temp.setPrev(cursor); temp.setNext(cursor.getNext()); cursor.getNext().setPrev(temp); cursor.setNext(temp); length++; } public static void main(String[] args) { int i = 0; i = getRandomNumber(10, 10000); DoublyLinkedList list = new DoublyLinkedList(); for (int j = 0; j < i; j++) { list.add(j, getRandomNumber(10, 10000)); MergeSort merge = new MergeSort(list); merge.sort(); System.out.println(list.getValue(j)); } } PLEASE help me thanks alot.

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  • Losing data after reading them correct from file

    - by user1388172
    i have the fallowing class of object with a class a data structure which i use in main combined. The ADT(abstract data type) is a linked list. After i read from file the input data and create and object which at print looks just fine after a print. after i push_back() the 3-rd int variable get initializated to 0. So example and code: Example: ex.in: 1 7 31 2 2 2 3 3 3 now i create objects from each line, which at print look as they suppose, but after push_back(): 1 7 0 2 2 0 3 3 0 Class.h: class RAngle { private: int x,y,l,b; public: int solution,prec; RAngle(){ x = y = solution = prec = b = l =0; } RAngle(int i,int j,int k){ x = i; y = j; l = k; solution = 0; prec=0; b=0; } friend ostream& operator << (ostream& out, const RAngle& ra){ out << ra.x << " " << ra.y << " " << ra.l <<endl; return out; } friend istream& operator >>( istream& is, RAngle& ra){ is >> ra.x; is >> ra.y; is >> ra.l; return is ; } }; ADT.h: template <class T> class List { private: struct Elem { T data; Elem* next; }; Elem* first; T pop_front(){ if (first!=NULL) { T aux = first->data; first = first->next; return aux; } T a; return a; } void push_back(T data){ Elem *n = new Elem; n->data = data; n->next = NULL; if (first == NULL) { first = n; return ; } Elem *current; for(current=first;current->next != NULL;current=current->next); current->next = n; } Main.cpp(after i call this function in main which prints object as they suppose to be the x var(from RAngle class) changes to 0 in all cases.) void readData(List <RAngle> &l){ RAngle r; ifstream f_in; f_in.open("ex.in",ios::in); for(int i=0;i<10;++i){ f_in >> r; cout << r; l.push_back(r); }

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  • How do I get Firefox to launch Visio when I click on a linked .vsd file?

    - by Dean
    On our intranet site, we have various MS Office documents linked. When I click on a Word, Excel or PowerPoint file, Firefox gives me the option to Open, Save or Cancel. When I click on Open, the appropriate app is launched and the file is loaded. This is perfect. But for some reason, when I click on a linked Visio file, I only get the option to Save, which is inconvenient. I know that Firefox knows the linked file is a Visio file because it tells me so in the dialog box: "You have chosen to open example.vsd which is a: Microsoft Visio Drawing". How can I make Firefox launch Visio when I click on a linked Visio file? Update: Firefox is not launching Visio when I click on a linked Visio file because the web server does not identify the content-type correctly. It identifies the Visio file as application/octet-stream instead of application/x-visio. (Thanks Grant Wagner.) This explains why it doesn't work. And in my case, I may be able to get the Apache config file changed, but this is not certain. However, I would love to know if there is a way to configure Firefox itself to launch Visio based on some other criteria, like file name extension. This way I can open Visio files even if I don't have access to the Apache configuration.

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  • Alias a linked Server in SQL server management studio?

    - by absentmindeduk
    Hoping someone can help - is there a way in SQL server management studio 2008 R2 that I can alias a linked SQL server? I have a server, added by IP address, to which I do not have the login credentials - however as the connection is already setup I can login ok. Issue is that, this is a dev environment, prior to a live deployment and the IP I have as a linked server needs to be 'accessible' by my stored procs under a different name, eg 'myserver' not 192.168.xxx.xxx... Any help much appreciated.

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  • Is this the right strategy to convert an in-level order binary tree to a doubly linked list?

    - by Ankit Soni
    So I recently came across this question - Make a function that converts a in-level-order binary tree into a doubly linked list. Apparently, it's a common interview question. This is the strategy I had - Simply do a pre-order traversal of the tree, and instead of returning a node, return a list of nodes, in the order in which you traverse them. i.e return a list, and append the current node to the list at each point. For the base case, return the node itself when you are at a leaf. so you would say left = recursive_function(node.left) right = recursive_function(node.right) return(left.append(node.data)).append(right);` Is this the right approach?

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  • Link List Problem,

    - by david
    OK i have a problem with a Link List program i'm trying to Do, the link List is working fine. Here is my code #include <iostream> using namespace std; struct record { string word; struct record * link; }; typedef struct record node; node * insert_Node( node * head, node * previous, string key ); node * search( node *head, string key, int *found); void displayList(node *head); node * delete_node( node *head, node * previous, string key); int main() { node * previous, * head = NULL; int found = 0; string node1Data,newNodeData, nextData,lastData; //set up first node cout <<"Depature"<<endl; cin >>node1Data; previous = search( head, node1Data, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, node1Data); cout <<"Depature inserted"<<endl; //insert node between first node and head cout <<"Destination"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Destinationinserted"<<endl; //insert node between second node and head cout <<"Cost"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Cost inserted"<<endl; cout <<"Number of Seats Required"<<endl; //place node between new node and first node cin >>nextData; previous = search( head, nextData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, nextData); cout <<"Number of Seats Required inserted"<<endl; //insert node between first node and head cout <<"Name"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Name inserted"<<endl; //insert node between node and head cout <<"Address "<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Address inserted"<<endl; //place node as very last node cin >>lastData; previous = search( head, lastData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, lastData); cout <<"C"<<endl; displayList(head); char Ans = 'y'; //Delete nodes do { cout <<"Enter Keyword to be delete"<<endl; cin >>nextData; previous = search( head, nextData, &found); if (found == 1) head = delete_node( head, previous,nextData); displayList(head); cout <<"Do you want to Delete more y /n "<<endl; cin >> Ans; } while( Ans =='y'); int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); } /******************************************************* search function to return previous position of node ******************************************************/ node * search( node *head, string key, int *found) { node * previous, * current; current = head; previous = current; *found = 0;//not found //if (current->word < key) move through links until the next link //matches or current_word > key while( current !=NULL) { //compare exactly if (key ==current->word ) { *found = 1; break; } //if key is less than word else if ( key < current->word ) break; else { //previous stays one link behind current previous = current; current = previous -> link; } } return previous; } /******************************************************** display function as used with createList ******************************************************/ void displayList(node *head) { node * current; //current now contains the address held of the 1st node similar //to head current = head; cout << "\n\n"; if( current ==NULL) cout << "Empty List\n\n"; else { /*Keep going displaying the contents of the list and set current to the address of the next node. When set to null, there are no more nodes */ while(current !=NULL) { cout << current->word<<endl; current = current ->link; } } } /************************************************************ insert node used to position node (i) empty list head = NULL (ii) to position node before the first node key < head->word (iii) every other position including the end of the list This is done using the following steps (a) Pass in all the details to create the node either details or a whole record (b) Pass the details over to fill the node (C) Use the if statement to add the node to the list **********************************************************/ node * insert_Node( node * head, node * previous, string key ) { node * new_node, * temp; new_node = new node; //create the node new_node ->word = key; new_node -> link = NULL; if (head == NULL || key < head->word ) //empty list { //give address of head to temp temp = head; //head now points to the new_node head = new_node; //new_node now points to what head was pointing at new_node -> link = temp; } else { //pass address held in link to temp temp = previous-> link; //put address of new node to link of previous previous -> link = new_node; //pass address of temp to link of new node new_node -> link = temp; } return head; } node * delete_node( node *head, node * previous, string key) { /* this function will delete a node but will not return its contents */ node * temp; if(key == head->word) //delete node at head of list { temp = head; //point head at the next node head = head -> link; } else { //holds the address of the node after the one // to be deleted temp = previous-> link; /*assign the previous to the address of the present node to be deleted which holds the address of the next node */ previous-> link = previous-> link-> link; } delete temp; return head; }//end delete The problem i have is when i Enter in the Number 2 in the Node(Seats) i like to get a Counter Taken 2 off of 50, some thing like what i have here enter code here int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); How can i get what the User enters into number of Seats into this function

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  • sp_addlinkserver using trigger

    - by Nanda
    I have the following trigger, which causes an error when it runs: CREATE TRIGGER ... ON ... FOR INSERT, UPDATE AS IF UPDATE(STATUS) BEGIN DECLARE @newPrice VARCHAR(50) DECLARE @FILENAME VARCHAR(50) DECLARE @server VARCHAR(50) DECLARE @provider VARCHAR(50) DECLARE @datasrc VARCHAR(50) DECLARE @location VARCHAR(50) DECLARE @provstr VARCHAR(50) DECLARE @catalog VARCHAR(50) DECLARE @DBNAME VARCHAR(50) SET @server=xx SET @provider=xx SET @datasrc=xx SET @provstr='DRIVER={SQL Server};SERVER=xxxxxxxx;UID=xx;PWD=xx;' SET @DBNAME='[xx]' SET @newPrice = (SELECT STATUS FROM Inserted) SET @FILENAME = (SELECT INPUT_XML_FILE_NAME FROM Inserted) IF @newPrice = 'FAIL' BEGIN EXEC master.dbo.sp_addlinkedserver @server, '', @provider, @datasrc, @provstr EXEC master.dbo.sp_addlinkedsrvlogin @server, 'true' INSERT INTO [@server].[@DBNAME].[dbo].[maildetails] ( 'to', 'cc', 'from', 'subject', 'body', 'status', 'Attachment', 'APPLICATION', 'ID', 'Timestamp', 'AttachmentName' ) VALUES ( 'P23741', '', '', 'XMLFAILED', @FILENAME, '4', '', '8', '', GETDATE(), '' ) EXEC sp_dropserver @server END END The error is: Msg 15002, Level 16, State 1, Procedure sp_MSaddserver_internal, Line 28 The procedure 'sys.sp_addlinkedserver' cannot be executed within a transaction. Msg 15002, Level 16, State 1, Procedure sp_addlinkedsrvlogin, Line 17 The procedure 'sys.sp_addlinkedsrvlogin' cannot be executed within a transaction. Msg 15002, Level 16, State 1, Procedure sp_dropserver, Line 12 The procedure 'sys.sp_dropserver' cannot be executed within a transaction. How can I prevent this error from occurring?

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  • How to reference a sql server with a slash (\) in its name?

    - by Bill Paetzke
    Givens: One SQL Server is named: DevServerA Another is named: DevServerB\2K5 Problem: From DevServerA, how can I write a query that references DevServerB\2K5? I tried a sample, dummy query (running it from DevServerA): SELECT TOP 1 * FROM DevServerB\2K5.master.sys.tables And I get the error: Msg 102, Level 15, State 1, Line 2 Incorrect syntax near '\.'. However, I know my syntax is almost correct, since the other way around works (running this query from DevServerB\2K5): SELECT TOP 1 * FROM DevServerA.master.sys.tables Please help me figure out how to reference DevServerB\2K5 from DevServerA. Thanks.

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  • printing reverse in singly link list using pointers

    - by theoneabhinav
    i have been trying this code. i think the logic is ok but the program terminates abruptly when the display_rev function is called here is code of display_rev void display_rev(emp_node *head) { emp_node *p=head, *q; while(p->next != NULL) p=p->next; while(p!=head || p==head){ q=head; display_rec(p); while(q->next != p) q=q->next; p=q; } } here is my whole code #include<stdio.h> #include<stdlib.h> #include<ctype.h> #include<string.h> //Declarations=============================================================== typedef struct //employee record { int emp_id; char name[150]; char mob_no[11]; float salary; int proj[5]; struct emp_node *next; } emp_node; emp_node* add_rec(emp_node*); emp_node* create_db(emp_node*); emp_node* search_db(emp_node*, int); emp_node* delete_rec(emp_node*, int); void read_name(emp_node*); void read_mob(emp_node*); void display_db(emp_node*); void display_rec(emp_node*); void display_rev(emp_node*); void modify_rec(emp_node*); void swtch(emp_node*); //=========================================================================== int main() { char ans; emp_node *head = NULL; head = create_db(head); display_db(head); do { swtch(head); printf("\n\n\tDo you want to continue (y/n) : "); getchar(); scanf("%c", &ans); } while (ans == 'y' || ans == 'Y'); return 0; } //Definitions================================================================ emp_node* create_db(emp_node *head) //database creation { int i = 1, no; emp_node *p; printf("Enter number of employees:"); scanf("%d", &no); printf("\n\n"); head = (emp_node *) malloc(sizeof(emp_node)); head = add_rec(head); head->next = NULL; p = head; while (i < no) { p->next = (emp_node *) malloc(sizeof(emp_node)); p = p->next; p = add_rec(p); p->next = NULL; i++; } return head; } emp_node* add_rec(emp_node *p) //new record { int j; printf("\n\tEmployee ID : "); scanf("%d", &(p->emp_id)); printf("\n\tFirst Name:"); read_name(p); printf("\n\tMobile No.:"); read_mob(p); printf("\n\tSalary :"); scanf("%f", &(p->salary)); printf( "\n\tEnter \"1\" for the projects employee is working on, otherwise enter \"0\": \n"); for (j = 0; j < 5; j++) { printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); while (p->proj[j] != 1 && p->proj[j] != 0) { printf("\n\nInvalid entry!! Please re-enter."); printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); } } printf("\n\n\n"); return p; } void read_name(emp_node *p) //validation for name { int j, len; scanf("%s", p->name); len = strlen(p->name); for (j = 0; j < len; j++) { if (!isalpha(p->name[j])) { printf( "\n\n\tInvalid name!!Can contain only characters. Please Re-enter.\n"); printf("\n\tName : "); read_name(p); } } } void read_mob(emp_node *p) //validation for mobile no. { int j; scanf("%s", p->mob_no); while (strlen(p->mob_no) != 10) { printf("\n\nInvalid Mobile No!!Please Re-enter"); printf("\n\n\tMobile No.:"); read_mob(p); } for (j = 0; j < 10; j++) { if (!(48 <= p->mob_no[j] && p->mob_no[j] <= 57)) { printf( "\n\nInvalid Mobile No!!Can contain only digits. Please Re-enter."); printf("\n\n\tMobile No.:"); read_mob(p); } } } void display_db(emp_node *head) //displaying whole database { emp_node *p; p = head; printf("\n\n\t\t****** EMPLOYEE DATABASE ******\n"); printf( "\n=============================================================================="); printf("\n Id.\t Name\t\t Mobile No\t Salary\t Projects\n"); while (p != NULL) { display_rec(p); p = p->next; printf("\n\n\n"); } printf( "\n=============================================================================="); } void swtch(emp_node *head) //function for menu and switch case { int cho, x; emp_node *p; printf("\n\n\t\t****** MENU ******"); printf( "\n\n\t1. insert Record\n\t2. Search Record\n\t3. Modify Record\n\t4. Delete Record\n\t5. Display Reverse\n\t6. Exit"); printf("\n\tWhich operation do you want to perform? "); scanf("%d", &cho); switch (cho) { case 1: p=head; while(p->next != NULL) p=p->next; p->next = (emp_node *) malloc(sizeof(emp_node)); p=p->next; p = add_rec(p); p->next = NULL; display_db(head); break; case 2: printf("\n\n\tEnter employee ID whose record is to be Searched :"); scanf("%d", &x); p = search_db(head, x); if (p == NULL) printf("\n\nRecord not found."); else display_rec(p); break; case 3: printf("\n\n\tEnter employee ID whose record is to be modified :"); scanf("%d", &x); p = search_db(head, x); if (p == NULL) printf("\n\nRecord not found."); else modify_rec(p); display_db(head); break; case 4: printf("\n\n\tEnter employee ID whose record is to be deleted :"); scanf("%d", &x); head = delete_rec(head, x); display_db(head); break; case 5: display_rev(head); case 6: exit(0); default: printf("Invalid Choice!! Please try again."); } } emp_node* search_db(emp_node *head, int id) //search database { emp_node *p = head; while (p != NULL) { if (p->emp_id == id) return p; p = p->next; } return NULL; } void display_rec(emp_node *p) //display a single record { int j; printf("\n %d", p->emp_id); printf("\t %10s", p->name); printf("\t %10s", p->mob_no); printf("\t %05.2f", p->salary); printf("\t "); for (j = 0; j < 5; j++) { if (p->proj[j] == 1) printf(" %d,", j + 1); } } void modify_rec(emp_node *p) //modifying a record { int j, cho; char ch1, edt; do { printf( "\n\t1. Name\n\t2. Email Address\n\t3. Mobile No.\n\t4. Salary\n\t5. Date of birth\n\t6. Projects\n"); printf("Enter your choice : "); scanf("%d", &cho); switch (cho) { case 1: printf("\n\tPrevious name:%s", p->name); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New Name:"); read_name(p); } break; case 2: printf("\n\tPrevious Mobile No. : %s", p->mob_no); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New Mobile No. :"); read_mob(p); } break; case 3: printf("\n\tPrevious salary is : %f", p->salary); printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf("\n\tEnter New salary:"); scanf("%f", &(p->salary)); } break; case 4: printf("the employee is currently working on project no. "); for (j = 0; j < 5; j++) { if (p->proj[j] == 1) printf(" %d,", j + 1); } printf("\n\tDo you want to edit ? (y/n)"); getchar(); scanf("%c", &ch1); if (ch1 == 'y' || ch1 == 'Y') { printf( "\n\tEnter \"1\" for the projects employee is working on : \n"); for (j = 0; j < 5; j++) { printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); while (p->proj[j] != 1) { printf("\n\nInvalid entry!! Please re-enter."); printf("\n\t\tProject No. %d : ", j + 1); scanf("%d", &(p->proj[j])); } } } break; default: printf("\n\nInvalid Choice!! Please Try again."); } printf("\n\nDo you want to edit any other fields ?(y/n)"); getchar(); scanf("%c", &edt); } while (edt == 'y' || edt == 'Y'); } emp_node* delete_rec(emp_node *head, int id) //physical deletion of record { emp_node *p = head, *q; if (head->emp_id == id) { head = head->next; free(p); return head; } else { q = p->next; while (q->emp_id != id) { p = p->next; q = q->next; } if (q->next == NULL) p->next = NULL; else p->next = q->next; free(q); return head; } } void display_rev(emp_node *head) { emp_node *p=head, *q; while(p->next != NULL) p=p->next; while(p!=head || p==head){ q=head; display_rec(p); while(q->next != p) q=q->next; p=q; } }

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  • Recursion Question : Revision

    - by stan
    My slides say that: A recursive call should always be on a smaller data structure than the current one There must be a non recursive option if the data structure is too small You need a wrapper method to make the recursive method accessible Just reading this from the slides makes no sense, especially seeing as it was a topic from before christmas! Could anyone try and clear up what it means please? Thank you

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  • Problems with sp_addlinkedserver in MSSQL 2000

    - by manneorama
    Hi! I'm having a bit of a problem with moving specific data from one server running MSSQL 2000 (A) and another running MSSQL 2008 (B). I'm writing a script according to customer specifications which is to be executed on A, populating tables in B with data. However, I can't seem to get the server link to work. -- Bunch of declarations here EXEC sp_addlinkedserver @server = @ServerName, @srvproduct = @ServerProduct, @provider = @ProviderString, @datasrc = @RemoteHost -- Data migration stuff here EXEC sp_dropserver @ServerName Now if I run the script in its entirety I get an error saying: Msg 7202, Level 11, State 2, Line 55 Could not find server 'remoteServer' in sysservers. Execute sp_addlinkedserver to add the server to sysservers. However, if I higlight only the sp_addlinkedserver-part and execute that, there is no error and I can highlight the rest of the script and run it. What am I missing here? Please help! PS. If backup-restore was an option, I would have done that already.

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  • Two seperate tm structs mirroring each other

    - by BSchlinker
    Here is my current situation: I have two tm structs, both set to the current time I make a change to the hour in one of the structs The change is occurring in the other struct magically.... How do I prevent this from occurring? I need to be able to compare and know the number of seconds between two different times -- the current time and a time in the future. I've been using difftime and mktime to determine this. I recognize that I don't technically need two tm structs (the other struct could just be a time_t loaded with raw time) but I'm still interested in understanding why this occurs. void Tracker::monitor(char* buffer){ // time handling time_t systemtime, scheduletime, currenttime; struct tm * dispatchtime; struct tm * uiuctime; double remainingtime; // let's get two structs operating with current time dispatchtime = dispatchtime_tm(); uiuctime = uiuctime_tm(); // set the scheduled parameters dispatchtime->tm_hour = 5; dispatchtime->tm_min = 05; dispatchtime->tm_sec = 14; uiuctime->tm_hour = 0; // both of these will now print the same time! (0:05:14) // what's linking them?? // print the scheduled time printf ("Current Time : %2d:%02d:%02d\n", uiuctime->tm_hour, uiuctime->tm_min, uiuctime->tm_sec); printf ("Scheduled Time : %2d:%02d:%02d\n", dispatchtime->tm_hour, dispatchtime->tm_min, dispatchtime->tm_sec); } struct tm* Tracker::uiuctime_tm(){ time_t uiucTime; struct tm *ts_uiuc; // give currentTime the current time time(&uiucTime); // change the time zone to UIUC putenv("TZ=CST6CDT"); tzset(); // get the localtime for the tz selected ts_uiuc = localtime(&uiucTime); // set back the current timezone unsetenv("TZ"); tzset(); // set back our results return ts_uiuc; } struct tm* Tracker::dispatchtime_tm(){ time_t currentTime; struct tm *ts_dispatch; // give currentTime the current time time(&currentTime); // get the localtime for the tz selected ts_dispatch = localtime(&currentTime); // set back our results return ts_dispatch; }

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  • How can I prevent segmentation faults in my program?

    - by worlds-apart89
    I have a C assignment. It is a lot longer than the code shown below, and we are given the function prototypes and instructions only. I have done my best at writing code, but I am stuck with segmentation faults. When I compile and run the program below on Linux, at "735 NaN" it will terminate, indicating a segfault occurred. Why? What am I doing wrong? Basically, the program does not let me access table-list_array[735]-value and table-list_array[735]-key. This is of course the first segfault. There might be more following index 735. #include <stdio.h> #include <stdlib.h> typedef struct list_node list_node_t; struct list_node { char *key; int value; list_node_t *next; }; typedef struct count_table count_table_t; struct count_table { int size; list_node_t **list_array; }; count_table_t* table_allocate(int size) { count_table_t *ptr = malloc(sizeof(count_table_t)); ptr->size = size; list_node_t *nodes[size]; int k; for(k=0; k<size; k++){ nodes[k] = NULL; } ptr->list_array = nodes; return ptr; } void table_addvalue(count_table_t *table) { int i; for(i=0; i<table->size; i++) { table->list_array[i] = malloc(sizeof(list_node_t)); table->list_array[i]->value = i; table->list_array[i]->key = "NaN"; table->list_array[i]->next = NULL; } } int main() { count_table_t *table = table_allocate(1000); table_addvalue(table); int i; for(i=0; i<table->size; i++) printf("%d %s\n", table->list_array[i]->value, table->list_array[i]->key); return 0; }

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  • Java threads not working correctly with linkedlist

    - by user69514
    Hi I am working on the sleeping barber problem. with the addition of having priority customer when they arrive they go in the front of the line and they are the next ones to get a haircut. I'm using a linkedlist and if I see a priority customer I put him in the beginning of the list, if the customer is not priority he goes to the end of the list. then I call the wantHaircut method getting the first element of the list. my problem is that the customer are being processed in the order they arrive, and the priority customer have to wait. here is the code where it all happens. any ideas what I am doing wrong? thanks public void arrivedBarbershop(Customer c){ if(waiting < numChairs && c.isPriority()){ System.out.println("Customer " + c.getID() + ": is a priority customer - SITTING -"); mutex.up(); customer_list.addFirst(c); } else if(waiting >= numChairs && c.isPriority()){ System.out.println("Customer " + c.getID() + ": is a priority customer - STANDING -"); mutex.up(); customer_list.addFirst(c); } else if(waiting < numChairs && !c.isPriority()){ waiting++; System.out.println("Customer " + c.getID() + ": arrived, sitting in the waiting room"); customer_list.addLast(c); customers.up(); // increment waiting customers } else if(waiting >= numChairs && !c.isPriority()) { System.out.println("Customer " + c.getID() + ": went to another barber because waiting room was full - " + waiting + " waiting"); mutex.up(); } if(!customer_list.isEmpty()){ this.wantHairCut(customer_list.removeFirst()); } } public void wantHairCut(Customer c) { mutex.up(); barber.down(); // waits for being allowed in barber chair System.out.println("Customer " + c.getID() + ": getting haircut"); try { /** haircut takes between 1 and 2 seconds **/ Thread.sleep(Barbershop.randomInt(1, 2) * 1000); } catch (InterruptedException e) { } System.out.println("Barber: finished cutting customer " + c.getID() + "'s hair"); c.gotHaircut = true; cutting.up(); // signals cutting has finished /** customer must pay now **/ this.wantToCashout(c); }

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  • Java - PriorityQueue vs sorted LinkedList

    - by msr
    Hello, Which implementation is less "heavy": PriorityQueue or a sorted LinkedList (using a Comparator)? I want to have all the items sorted. The insertion will be very frequent and ocasionally I will have to run all the list to make some operations. Thank you!

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  • When to use LinkedList<> over ArrayList<>?

    - by sdellysse
    I've always been one to simply use List<String> names = new ArrayList<String>(); I use the interface as the type name for portability, so that when I ask questions such as these I can rework my code. When should LinkedList should be used over ArrayList and vice-versa?

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  • SQL Server Error: maximum number of prefixes. The maximum is 3.

    - by Ian Boyd
    Trying to run a cross-server update: UPDATE cmslive.CMSFintrac.dbo.lsipos SET PostHistorySequencenNmber = ( SELECT TransactionNumber FROM Transactions WHERE Transactions.TransactionDate = cmslive.CMSFintrac.dbo.lsipos.TransactionDate) Gives the error: Server: Msg 117, Level 15, State 2, Line 5 The number name 'cmslive.CMSFintrac.dbo.lsipos' contains more than the maximum number of prefixes. The maximum is 3. What gives? Note: Rearranging the query into a less readable join form: UPDATE cmslive.CMSFintrac.dbo.lsipos SET PostHistorySequenceNumber = B.TransactionNumber FROM cmslive.CMSFintrac.dbo.lsipos A INNER JOIN Transactions B ON A.TransactionDate = B.TransactionDate does not give an error.

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  • Clone List Elements in Java

    - by Amir Rachum
    Hi all, I have a variable of type List<RelationHeader>. Now I want to copy all the elements in this list to a new list, but I want to actually copy all the members by value (clone them). Is there a quick command to do this, or do I need to iterate over the list and copy them one at a time?

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  • How come Java doesn't accept my LinkedList in a Generic, but accepts its own?

    - by master chief
    For a class assignment, we can't use any of the languages bultin types, so I'm stuck with my own list. Anyway, here's the situation: public class CrazyStructure <T extends Comparable<? super T>> { MyLinkedList<MyTree<T>> trees; //error: type parameter MyTree is not within its bound } However: public class CrazyStructure <T extends Comparable<? super T>> { LinkedList<MyTree<T>> trees; } Works. MyTree impleements the Comparable interface, but MyLinkedList doesn't. However, Java's LinkedList doesn't implement it either, according to this. So what's the problem and how do I fix it? MyLinkedList: public class MyLinkedList<T extends Comparable<? super T>> { private class Node<T> { private Node<T> next; private T data; protected Node(); protected Node(final T value); } Node<T> firstNode; public MyLinkedList(); public MyLinkedList(T value); //calls node1.value.compareTo(node2.value) private int compareElements(final Node<T> node1, final Node<T> node2); public void insert(T value); public void remove(T value); } MyTree: public class LeftistTree<T extends Comparable<? super T>> implements Comparable { private class Node<T> { private Node<T> left, right; private T data; private int dist; protected Node(); protected Node(final T value); } private Node<T> root; public LeftistTree(); public LeftistTree(final T value); public Node getRoot(); //calls node1.value.compareTo(node2.value) private int compareElements(final Node node1, final Node node2); private Node<T> merge(Node node1, Node node2); public void insert(final T value); public T extractMin(); public int compareTo(final Object param); }

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  • LinkedList insert tied to inserted object

    - by wrongusername
    I have code that looks like this: public class Polynomial { List<Term> term = new LinkedList<Term>(); and it seems that whenever I do something like term.add(anotherTerm), with anotherTerm being... another Term object, it seems anotherTerm is referencing the same thing as what I've just inserted into term so that whenever I try to change anotherTerm, term.get(2) (let's say) get's changed too. How can I prevent this from happening?

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  • Recurssion Question : Revision

    - by stan
    My slides say that A recurssive call should always be on a smaller data structure than the current one There must be a non recurssive option if the data structure is too small You need a wrapper method to make the recurssive method accessible Just reading this from the slides makes no sence, especially seeing as it was a topic from before christmas! Could anyone try and clear up what it means please? Thank you

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