Search Results

Search found 45600 results on 1824 pages for 'top list'.

Page 31/1824 | < Previous Page | 27 28 29 30 31 32 33 34 35 36 37 38  | Next Page >

  • pointer to a pointer in a linked list

    - by user1596497
    I'm trying to set a linked list head through pointer to a pointer. I can see inside the function that the address of the head pointer is changing but as i return to the main progran it becomes NULL again. can someone tell me what I'm doing wrong ?? #include <stdio.h> #include <stdlib.h> typedef void(*fun_t)(int); typedef struct timer_t { int time; fun_t func; struct timer_t *next; }TIMER_T; void add_timer(int sec, fun_t func, TIMER_T *head); void run_timers(TIMER_T **head); void timer_func(int); int main(void) { TIMER_T *head = NULL; int time = 1; fun_t func = timer_func; while (time < 1000) { printf("\nCalling add_timer(time=%d, func=0x%x, head=0x%x)\n", time, func, &head); add_timer(time, func, head); time *= 2; } run_timers(&head); return 0; } void add_timer(int sec, fun_t func, TIMER_T *head) { TIMER_T ** ppScan=&head; TIMER_T *new_timer = NULL; new_timer = (TIMER_T*)malloc(sizeof(TIMER_T)); new_timer->time = sec; new_timer->func = func; new_timer->next = NULL; while((*ppScan != NULL) && (((**ppScan).time)<sec)) ppScan = &(*ppScan)->next; new_timer->next = *ppScan; *ppScan = new_timer; }

    Read the article

  • Should xml represent a set or a list?

    - by sixtyfootersdude
    I always think of xml like a set data structure. Ie: <class> <person>john</person> <person>sarah</person> </class> Is equivalent to: <class> <person>sarah</person> <person>john</person> </class> Question One: Are these two things logicly equivalant? Are you allowed to make things like this in xml? <methodCall> <param>happy</param> <param>sad</param> </methodCall> Or do you need to do it like this: <methodCall> <param arg="1">happy</param> <param arg="2">sad</param> </methodCall> Question Two: Are these two things logically equivalent? Question Three: Is xml usually treated like a set or a list?

    Read the article

  • adding elements in to the doubly linked list

    - by user329820
    Hi this is my code for main class and doubly linked class and node class but when I run the program ,in the concole will show this"datastructureproject.DoublyLinkedList@19ee1ac" instead of the random numbers .please help me thanks! main class: public class Main { public static int getRandomNumber(double min, double max) { Random random = new Random(); return (int) (random.nextDouble() * (max - min) + min); } public static void main(String[] args) { int j; int i = 0; i = getRandomNumber(10, 10000); DoublyLinkedList listOne = new DoublyLinkedList(); for (j = 0; j <= i / 2; j++) { listOne.add(getRandomNumber(10, 10000)); } System.out.println(listOne); } } doubly linked list class: public class DoublyLinkedList { private Node head ; private Node tail; private long size = 0; public DoublyLinkedList() { head= new Node(0, null, null); tail = new Node(0, head, null); } public void add(int i){ head.setValue(i); Node newNode = new Node(); head.setNext(newNode); newNode.setPrev(head); newNode = head; } } and the node class is like the class that you have seen before (Node prev,Node next,int value)

    Read the article

  • Linked List Inserting strings in alphabetical order

    - by user69514
    I have a linked list where each node contains a string and a count. my insert method needs to inset a new node in alphabetical order based on the string. if there is a node with the same string, then i increment the count. the problem is that my method is not inserting in alphabetical order public Node findIsertionPoint(Node head, Node node){ if( head == null) return null; Node curr = head; while( curr != null){ if( curr.getValue().compareTo(node.getValue()) == 0) return curr; else if( curr.getNext() == null || curr.getNext().getValue().compareTo(node.getValue()) > 0) return curr; else curr = curr.getNext(); } return null; } public void insert(Node node){ Node newNode = node; Node insertPoint = this.findIsertionPoint(this.head, node); if( insertPoint == null) this.head = newNode; else{ if( insertPoint.getValue().compareTo(node.getValue()) == 0) insertPoint.getItem().incrementCount(); else{ newNode.setNext(insertPoint.getNext()); insertPoint.setNext(newNode); } } count++; }

    Read the article

  • Deleting first and last element of a linked list in C

    - by LuckySlevin
    struct person { int age; char name[100]; struct person *next; }; void delfirst(struct person **p)// For deleting the beginning { struct person *tmp,*m; m = (*p); tmp = (*p)->next; free(m); return; } void delend(struct person **p)// For deleting the end { struct person *tmp,*m; tmp=*p; while(tmp->next!=NULL) { tmp=tmp->next; } m->next=tmp; free(tmp); m->next = NULL; return; } I'm looking for two seperate functions to delete the first and last elements of a linked list. Here is what i tried. What do you suggest? Especially deleting first is so problematic for me.

    Read the article

  • Cleaner method for list comprehension clean-up

    - by Dan McGrath
    This relates to my previous question: Converting from nested lists to a delimited string I have an external service that sends data to us in a delimited string format. It is lists of items, up to 3 levels deep. Level 1 is delimited by '|'. Level 2 is delimited by ';' and level 3 is delimited by ','. Each level or element can have 0 or more items. An simplified example is: a,b;c,d|e||f,g|h;; We have a function that converts this to nested lists which is how it is manipulated in Python. def dyn_to_lists(dyn): return [[[c for c in b.split(',')] for b in a.split(';')] for a in dyn.split('|')] For the example above, this function results in the following: >>> dyn = "a,b;c,d|e||f,g|h;;" >>> print (dyn_to_lists(dyn)) [[['a', 'b'], ['c', 'd']], [['e']], [['']], [['f', 'g']], [['h'], [''], ['']]] For lists, at any level, with only one item, we want it as a scalar rather than a 1 item list. For lists that are empty, we want them as just an empty string. I've came up with this function, which does work: def dyn_to_min_lists(dyn): def compress(x): return "" if len(x) == 0 else x if len(x) != 1 else x[0] return compress([compress([compress([item for item in mv.split(',')]) for mv in attr.split(';')]) for attr in dyn.split('|')]) Using this function and using the example above, it returns: [[['a', 'b'], ['c', 'd']], 'e', '', ['f', 'g'], ['h', '', '']] Being new to Python, I'm not confident this is the best way to do it. Are there any cleaner ways to handle this? This will potentially have large amounts of data passing through it, are there any more efficient/scalable ways to achieve this?

    Read the article

  • Python - creating a list with 2 characteristics bug

    - by user2733911
    The goal is to create a list of 99 elements. All elements must be 1s or 0s. The first element must be a 1. There must be 7 1s in total. import random import math import time # constants determined through testing generation_constant = 0.96 def generate_candidate(): coin_vector = [] coin_vector.append(1) for i in range(0, 99): random_value = random.random() if (random_value > generation_constant): coin_vector.append(1) else: coin_vector.append(0) return coin_vector def validate_candidate(vector): vector_sum = sum(vector) sum_test = False if (vector_sum == 7): sum_test = True first_slot = vector[0] first_test = False if (first_slot == 1): first_test = True return (sum_test and first_test) vector1 = generate_candidate() while (validate_candidate(vector1) == False): vector1 = generate_candidate() print vector1, sum(vector1), validate_candidate(vector1) Most of the time, the output is correct, saying something like [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0] 7 True but sometimes, the output is: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 2 False What exactly am I doing wrong?

    Read the article

  • Best depth sorting method for a Top Down 2D game using a 3D physics engine

    - by Alic44
    I've spent many days googling this and still have issues with my game engine I'd like to ask about, which I haven't seen addressed before. I think the problem is that my game is an unusual combination of a completely 2D graphical approach using XNA's SpriteBatch, and a completely 3D engine (the amazing BEPU physics engine) with rotation mostly disabled. In essence, my question is similar to this one (the part about "faux 3D"), but the difference is that in my game, the player as well as every other creature is represented by 3D objects, and they can all jump, pick up other objects, and throw them around. What this means is that sorting by one value, such as a Z position (how far north/south a character is on the screen) won't work, because as soon as a smaller creature jumps on top of a larger creature, or a box, and walks backwards, the moment its z value is less than that other creature, it will appear to be behind the object it is actually standing on. I actually originally solved this problem by splitting every object in the game into physics boxes which MUST have a Y height equal to their Z depth. I then based the depth sorting value on the object's y position (how high it is off the ground) PLUS its z position (how far north or south it is on the screen). The problem with this approach is that it requires all moving objects in the game to be split graphically into chunks which match up with a physical box which has its y dimension equal to its z dimension. Which is stupid. So, I got inspired last night to rewrite with a fresh approach. My new method is a little more complex, but I think a little more sane: every object which needs to be sorted by depth in the game exposes the interface IDepthDrawable and is added to a list owned by the DepthDrawer object. IDepthDrawable contains: public interface IDepthDrawable { Rectangle Bounds { get; } //possibly change this to a class if struct copying of the xna Rectangle type becomes an issue DepthDrawShape DepthShape { get; } void Draw(SpriteBatch spriteBatch); } The Bounds Rectangle of each IDepthDrawable object represents the 2D Axis-Aligned Bounding Box it will take up when drawn to the screen. Anything that doesn't intersect the screen will be culled at this stage and the remaining on-screen IDepthDrawables will be Bounds tested for intersections with each other. This is where I get a little less sure of what I'm doing. Each group of collisions will be added to a list or other collection, and each list will sort itself based on its DepthShape property, which will have access to the object-to-be-drawn's physics information. For starting out, lets assume everything in the game is an axis aligned 3D Box shape. Boxes are pretty easy to sort. Something like: if (depthShape1.Back > depthShape2.Front) //if depthShape1 is in front of depthShape2. //depthShape1 goes on top. else if (depthShape1.Bottom > depthShape2.Top) //if depthShape1 is above depthShape2. //depthShape1 goes on top. //if neither of these are true, depthShape2 must be in front or above. So, by sorting draw order by several different factors from the physics engine, I believe I can get a really correct draw order. My question is, is this a good way of going about this, or is there some tried and true, tested way which is completely different and has somehow completely eluded me on the internets? And, if this does seem like a good way to remake my draw order sorting, what's the right sorting algorithm for reordering the Bounds Rectangle collision lists, and how do you deal with a Bounds Rectangle colliding with two different object which don't collide with eachother. I know these are solved problems, but I've only been programming for a year so any specific input here will be greatly appreciated. Thanks for reading this far, ye who made it -- sorry it was so long!

    Read the article

  • Simple Constructor With Initializer List? - C++

    - by Alex
    Hi all, below I've included my h file, and my problem is that the compiler is not liking my simple exception class's constructor's with initializer lists. It also is saying that string is undeclared identifier, even though I have #include <string> at the top of the h file. Do you see something I am doing wrong? For further explanation, this is one of my domain classes that I'm integrating into a wxWidgets GUI application on Windows. Thanks! Time.h #pragma once #include <string> #include <iostream> // global constants for use in calculation const int HOURS_TO_MINUTES = 60; const int MINUTES_TO_HOURS = 100; class Time { public: // default Time class constructor // initializes all vars to default values Time(void); // ComputeEndTime computes the new delivery end time // params - none // preconditions - vars will be error-free // postconditions - the correct end time will be returned as an int // returns an int int ComputeEndTime(); // GetStartTime is the getter for var startTime // params - none // returns an int int GetStartTime() { return startTime; } // GetEndTime is the getter for var endTime // params - none // returns an int int GetEndTime() { return endTime; } // GetTimeDiff is the getter for var timeDifference // params - none // returns a double double GetTimeDiff() { return timeDifference; } // SetStartTime is the setter for var startTime // params - an int // returns void void SetStartTime(int s) { startTime = s; } // SetEndTime is the setter for var endTime // params - an int // returns void void SetEndTime(int e) { endTime = e; } // SetTimeDiff is the setter for var timeDifference // params - a double // returns void void SetTimeDiff(double t) { timeDifference = t; } // destructor for Time class ~Time(void); private: int startTime; int endTime; double timeDifference; }; class HourOutOfRangeException { public: // param constructor // initializes message to passed paramater // preconditions - param will be a string // postconditions - message will be initialized // params a string // no return type HourOutOfRangeException(string pMessage) : message(pMessage) {} // GetMessage is getter for var message // params none // preconditions - none // postconditions - none // returns string string GetMessage() { return message; } // destructor ~HourOutOfRangeException() {} private: string message; }; class MinuteOutOfRangeException { public: // param constructor // initializes message to passed paramater // preconditions - param will be a string // postconditions - message will be initialized // params a string // no return type MinuteOutOfRangeException(string pMessage) : message(pMessage) {} // GetMessage is getter for var message // params none // preconditions - none // postconditions - none // returns string string GetMessage() { return message; } // destructor ~MinuteOutOfRangeException() {} private: string message; }; class PercentageOutOfRangeException { public: // param constructor // initializes message to passed paramater // preconditions - param will be a string // postconditions - message will be initialized // params a string // no return type PercentageOutOfRangeException(string pMessage) : message(pMessage) {} // GetMessage is getter for var message // params none // preconditions - none // postconditions - none // returns string string GetMessage() { return message; } // destructor ~PercentageOutOfRangeException() {} private: string message; }; class StartEndException { public: // param constructor // initializes message to passed paramater // preconditions - param will be a string // postconditions - message will be initialized // params a string // no return type StartEndException(string pMessage) : message(pMessage) {} // GetMessage is getter for var message // params none // preconditions - none // postconditions - none // returns string string GetMessage() { return message; } // destructor ~StartEndException() {} private: string message; };

    Read the article

  • Top 10 Browser Productivity Tips

    - by Renso
    Originally posted on: http://geekswithblogs.net/renso/archive/2013/10/14/top-10-browser-productivity-tips.aspxYou don’t have to be a geek to be a productive browser user. The tips below have been selected by actions users take most of the time to navigate a web-site but use long-standing keyboard or mouse actions to get them done, when there are keyboard short-cuts you can use instead. Since you hands are already on the keyboard it is almost always faster to sue a keyboard shortcut to get something done that you usually used the mouse for. For example right-clicking on something to copy it and then doing to same for pasting something is very time consuming, keyboard shortcuts have been created that simplify the task. All it takes are a few memory brain cells to remember them. Here are the tips, in no particular order:   Tip 1 Hold down the spacebar on your keyboard to page to the end of your web page rather than using your mouse. This is really a slow way of doing it. If you want to page one page at a time, hit the spacebar once, and again to page again. But if you want to page all the way to the end of the web page simply hit Ctrl+End (that is hold down the Ctrl key and hit the End button on your keyboard). To get to the top of your web page, simply hit Ctrl + Home to go all the way to the top of your web page. Tip 2 Where are my downloads? Some folks run downloads again-and-again because they do not know where the last one went and they do not see the popup, or browser note on their web page in the footer, etc. Simply hit Ctrl+J. Works in most browsers. Tip 3 Selecting a US state from a drop down box. Don’t use the mouse, takes just way too long to scroll. When you tab to the drop down box or click on it with your mouse, simply hit the first character of the state and it will be selected. For Texas for example hit the letter “T” twice on your keyboard to get to it. The same concept can be applied to any drop down box that is alphabetical or numerically sorted. Tip 4 Fixing spelling errors. All modern-day browsers support this now. You see the red wavy lines underscoring a word, yes it is a spelling error. How do you fix it? Don’t overtype it or try and fix it manually, fist right-click on it and a list of suggestions comes up. If it does not show up, like my name “Renso” and you know how to spell your name as in this example, look further down the list of options (the little window popup that appears when you right click) and you should see an option to “Add to Dictionary”. Be warned, when you add it, it only adds it to the browser you’re using’s dictionary. If you use Google Chrome, Firefox and IE, each one will have their own list. Tip 5 So you have trouble seeing the text on the screen. Or you are looking at a photo, for example in Facebook. You want to zoom in to read better or zoom into a photo a bit more. Hit Ctrl++ (hold down Ctrl key and hit the plus key – actually it’s the equal key but it is easier to remember that it is plus for bigger). Hit the minus to zoom out. Now you can’t remember what the original size was since you were so excited to hit it 20 times, or was that 21… Simply hit Ctrl+0 (that is zero) and it will reset it to the default. Tip 6 So you closed a couple of tabs in your browser. Suddenly you remember something you wanted to double-check something on one of the tabs, you cannot remember the URL ad the tab is gone forever, or is it? Simply hit Ctrl+Shift+t and it will bring back your tabs one by one each time you click the T. This has also been a great way for me to quickly close some tabs because I don’t want my boss to see I’m shopping and then hitting Ctrl+Shift+t to quickly get it back and complete my check-put and purchase. Or, for parents, when you walk into your daughter’s room and you see she quickly clicks and closes a window/tab in here browser. Not to worry my little darling, daddy will Ctrl+Shift+t and see what boys on Facebook you were talking too… Tip 7 The web browser is frozen on your PC/Laptop/Whatever, in this example it may be your Internet Explorer browser. I don’t mention Firefox or Chrome here because it probably never happens in their world. You cannot close it, it won’t respond to anything you have done s far except for the next step you are about to take, which is throw your two-day old coffee on your keyboard. This happens especially on sites that want to force you to complete a purchase order. Hit Ctrl+Alt+Del on your keyboard on any version of windows, select TASK MANAGER. In the  First Tab, which is the Process Tab, look for the item in question. In this example you should see Internet Explorer. Right-click it and select “End Task”. It will force the thread out of memory and terminate that process. You can of course do this with any program running under your account. Tip 8 This is a personal favorite of mine. To select words in the paragraph without using the mouse. You don’t want to select one character at a time like when you use the Ctrl+arrows as it can be very slow if you want to select a lot of text. You also want to select whole words. Simply use the Ctrl+Shift_arrow (right or left depending which direction you want to go. Tip 9 I was a bit reluctant to add this one, but being in the professional services industry still come across many-a-folk that simply can’t copy-and-paste them-all text or images that reside on them screens, y’all. Ctrl+c to copy and Ctrl+v to paste it. Works a lot faster than using the mouse. You may be asking: “Well why in the devil did they not use Ctrl+p for paste…. because that is for printing. This is of course not limited to the browser world, it applies to almost any piece of software running on PC or Mac. Go try it on an image on your browser, right-click it and select copy. Open a word document and Ctrl+v to paste the image in there. Please consider copyright laws. Tip 10 Getting rid of annoying ads. Now this only works when you load a web page, meaning when you get back to the same page later you will have to do this again and you will need to learn a tool to do it, WELL WORTH IT. For example, I use GrooveShark to listen to music but I don’t like the ads they show. Install a tool like Firebug for Firefox or use the Ctrl+Shift+I on Chrome to bring up the developer toolbar. Shows at the bottom of the page. With Firefox, once you have installed Firebug as an add-on, a yellow bug should appear on the top right-hand-side of your browser, click on it to display the developer toolbar. You will need to learn how to use it, but once you know how to select an item/section on the window (usually just right-click the add you don’t want to see and select “Inspect Element”, the developer toolbar will appear (if not already there)) and then simply hit delete and it will remove the add from the screen. If you don’t know HTML you may need to play with it a bit, but once you understand how it works can open up a whole new world for you on how web pages actually work. If you can think of any others that have saved you a ton of time please let me know so I can add them to a top 99 list.

    Read the article

  • print a linear linked list into a table

    - by user1796970
    I am attempting to print some values i have stored into a LLL into a readable table. The data i have stored is the following : DEBBIE STARR F 3 W 1000.00 JOAN JACOBUS F 9 W 925.00 DAVID RENN M 3 H 4.75 ALBERT CAHANA M 3 H 18.75 DOUGLAS SHEER M 5 W 250.00 SHARI BUCHMAN F 9 W 325.00 SARA JONES F 1 H 7.50 RICKY MOFSEN M 6 H 12.50 JEAN BRENNAN F 6 H 5.40 JAMIE MICHAELS F 8 W 150.00 i have stored each firstname, lastname, gender, tenure, payrate, and salary into their own List. And would like to be able to print them out in the same format that they are viewed on the text file i read them in from. i have messed around with a few methods that allow me to traverse and print the Lists, but i end up with ugly output. . . here is my code for the storage of the text file and the format i would like to print out: public class Payroll { private LineWriter lw; private ObjectList output; ListNode input; private ObjectList firstname, lastname, gender, tenure, rate, salary; public Payroll(LineWriter lw) { this.lw = lw; this.firstname = new ObjectList(); this.lastname = new ObjectList(); this.gender = new ObjectList(); this.tenure = new ObjectList(); this.rate = new ObjectList(); this.salary = new ObjectList(); this.output = new ObjectList(); this.input = new ListNode(); } public void readfile() { File file = new File("payfile.txt"); try{ Scanner scanner = new Scanner(file); while(scanner.hasNextLine()) { String line = scanner.nextLine(); Scanner lineScanner = new Scanner(line); lineScanner.useDelimiter("\\s+"); while(lineScanner.hasNext()) { firstname.insert1(lineScanner.next()); lastname.insert1(lineScanner.next()); gender.insert1(lineScanner.next()); tenure.insert1(lineScanner.next()); rate.insert1(lineScanner.next()); salary.insert1(lineScanner.next()); } } }catch(FileNotFoundException e) {e.printStackTrace();} } public void printer(LineWriter lw) { String msg = " FirstName " + " LastName " + " Gender " + " Tenure " + " Pay Rate " + " Salary "; output.insert1(msg); System.out.println(output.getFirst()); System.out.println(" " + firstname.getFirst() + " " + lastname.getFirst() + "\t" + gender.getFirst() + "\t" + tenure.getFirst() + "\t" + rate.getFirst() + "\t" + salary.getFirst()); } }

    Read the article

  • WPF, C# - Making Intellisense/Autocomplete list, fastest way to filter list of strings

    - by user559548
    Hello everyone, I'm writing an Intellisense/Autocomplete like the one you find in Visual Studio. It's all fine up until when the list contains probably 2000+ items. I'm using a simple LINQ statement for doing the filtering: var filterCollection = from s in listCollection where s.FilterValue.IndexOf(currentWord, StringComparison.OrdinalIgnoreCase) >= 0 orderby s.FilterValue select s; I then assign this collection to a WPF Listbox's ItemSource, and that's the end of it, works fine. Noting that, the Listbox is also virtualised as well, so there will only be at most 7-8 visual elements in memory and in the visual tree. However the caveat right now is that, when the user types extremely fast in the richtextbox, and on every key up I execute the filtering + binding, there's this semi-race condition, or out of sync filtering, like the first key stroke's filtering could still be doing it's filtering or binding work, while the fourth key stroke is also doing the same. I know I could put in a delay before applying the filter, but I'm trying to achieve a seamless filtering much like the one in Visual Studio. I'm not sure where my problem exactly lies, so I'm also attributing it to IndexOf's string operation, or perhaps my list of string's could be optimised in some kind of index, that could speed up searching. Any suggestions of code samples are much welcomed. Thanks.

    Read the article

  • Double Linked List Insertion Sorting Bug

    - by house
    Hello, I have implemented an insertion sort in a double link list (highest to lowest) from a file of 10,000 ints, and output to file in reverse order. To my knowledge I have implemented such a program, however I noticed in the ouput file, a single number is out of place. Every other number is in correct order. The number out of place is a repeated number, but the other repeats of this number are in correct order. Its just strange how this number is incorrectly placed. Also the unsorted number is only 6 places out of sync. I have looked through my program for days now with no idea where the problem lies, so I turn to you for help. Below is the code in question, (side note: can my question be deleted by myself? rather my colleges dont thieve my code, if not how can it be deleted?) void DLLIntStorage::insertBefore(int inValue, node *nodeB) { node *newNode; newNode = new node(); newNode->prev = nodeB->prev; newNode->next = nodeB; newNode->value = inValue; if(nodeB->prev==NULL) { this->front = newNode; } else { nodeB->prev->next = newNode; } nodeB->prev = newNode; } void DLLIntStorage::insertAfter(int inValue, node *nodeB) { node *newNode; newNode = new node(); newNode->next = nodeB->next; newNode->prev = nodeB; newNode->value = inValue; if(nodeB->next == NULL) { this->back = newNode; } else { nodeB->next->prev = newNode; } nodeB->next = newNode; } void DLLIntStorage::insertFront(int inValue) { node *newNode; if(this->front == NULL) { newNode = new node(); this->front = newNode; this->back = newNode; newNode->prev = NULL; newNode->next = NULL; newNode->value = inValue; } else { insertBefore(inValue, this->front); } } void DLLIntStorage::insertBack(int inValue) { if(this->back == NULL) { insertFront(inValue); } else { insertAfter(inValue, this->back); } } ifstream& operator>> (ifstream &in, DLLIntStorage &obj) { int readInt, counter = 0; while(!in.eof()) { if(counter==dataLength) //stops at 10,000 { break; } in >> readInt; if(obj.front != NULL ) { obj.insertion(readInt); } else { obj.insertBack(readInt); } counter++; } return in; } void DLLIntStorage::insertion(int inValue) { node* temp; temp = this->front; if(temp->value >= inValue) { insertFront(inValue); return; } else { while(temp->next!=NULL && temp!=this->back) { if(temp->value >= inValue) { insertBefore(inValue, temp); return; } temp = temp->next; } } if(temp == this->back) { insertBack(inValue); } } Thankyou for your time.

    Read the article

  • Linked list in C

    - by ScReYm0
    I am still new at lists in C and i got a big problem... First i wanna show you my code for inserting item to the list: void input_books_info(int number_of_books, BOOK *current) { int i; for(i = 0; i < number_of_books; i++) { while(current->next != NULL) current = current->next; current->next = (BOOK *)malloc(sizeof(BOOK)); printf_s("%d book catalog number: ", i + 1); scanf_s("%s", &current->next->catalog_number , 20); printf_s("%d book title: ", i + 1); scanf_s("%s", current->next->title ,80); printf_s("%d book author: ", i + 1); scanf_s("%s", current->next->author ,40); printf_s("%d book publisher: ", i+1); scanf_s("%s", current->next->publisher,80); printf_s("%d book price: ", i + 1); scanf_s("%f", &current->next->price, 5); printf_s("%d book year published: ", i + 1); scanf_s("%d", &current->next->year_published, 5); current->next->next = NULL; printf_s("\n\n"); } } And this is my main function: void main (void) { int number_of_books, t = 1; char book_catalog_number[STRMAX]; char book_title[STRMAX]; char book_author[STRMAX]; char reading_file[STRMAX]; char saving_file[STRMAX]; first = malloc(sizeof(BOOK)); first->next = NULL; /* printf_s("Enter file name: "); gets(saving_file); first->next = book_open(first, saving_file); */ while(t) { char m; printf_s("1. Input \n0. Exit \n\n"); printf_s("Choose operation: "); m = getch(); switch(m) { case '1': printf_s("\ninput number of books: "); scanf_s("%d", &number_of_books); input_books_info(number_of_books, first); printf_s("\n"); break; default: printf_s("\nNo entry found!\n\n\n\n\n"); break; } } } and last maybe here is the problem the printing function: void print_books_info(BOOK *current) { while(current->next != NULL && current != NULL) { printf_s("%s, ", current->next->catalog_number); printf_s("%s, ", current->author); printf_s("%s, ", current->next->title); printf_s("%s, ", current->next->author); printf_s("%s, ", current->next->publisher); printf_s("%.2f, ", current->next->price); printf_s("%d", current->next->year_published); printf_s("\n\n"); current = current->next; } } And my problem is that, when i run the app, program is moving good. But when I start the app, the program is storing only first input of data second and third are lost ... Can you help me to figure out it... ???

    Read the article

  • Unable to add separator in list view

    - by Suru
    This is my code @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.email_list_main); emailResults = new ArrayList<GetEmailFromDatabase>(); //int[] colors = {0,0xFFFF0000,0}; //getListView().setDivider(new GradientDrawable(Orientation.RIGHT_LEFT, colors)); //getListView().setDividerHeight(2); emailListFeedAdapter = new EmailListFeedAdapter(this, R.layout.email_listview_row, emailResults); setListAdapter(this.emailListFeedAdapter); getResults(); if(emailResults != null && emailResults.size() > -1){ emailListFeedAdapter.notifyDataSetChanged(); for(int i=0;i< emailResults.size();i++){ try { Here I getting email Sent date emailListFeedAdapter.add( emailResults.get(i)); datetime_text1 = emailResults.get(i).getDate(); formatter1 = new SimpleDateFormat(); formatter1 = DateFormat.getDateInstance((DateFormat.MEDIUM)); Calendar currentDate1 = Calendar.getInstance(); Item_Date1 = formatter1.parse(datetime_text1); current_Date1 = formatter1.format(currentDate1.getTime()); current_System_Date1 = formatter1.parse(current_Date1); currentDate1.add(Calendar.DATE, -1); yesterdaydate = formatter1.format(currentDate1.getTime()); yeaterday_Date = formatter1.parse(yesterdaydate); currentDate1.add(Calendar.DATE, -2); threeDaysback = formatter1.format(currentDate1.getTime()); Three_Days_Back = formatter1.parse(threeDaysback); Here I am comparing current date with list view item date, and here is my problem, dates are matching but it is not entering in if condition I tried in so many ways but nothing worked the code for separator is bellow. if(Item_Date.compareTo(current_System_Date)==0){ if(index1){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index1=false; } } else if(yeaterday_Date.compareTo(Item_Date1)==0){ if(index2){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index2 = false; } } else if(Item_Date1.compareTo(Three_Days_Back)==0){ if(index3){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index3 = false; } } } catch (ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } } } } In EmailListFeedAdapter private TreeSet<Integer> mSeparatorsSet = new TreeSet<Integer>(); public void addSeparatorItem(final String item) { //itemss.add(emailResults.get(0)); // save separator position mSeparatorsSet.add(itemss.size() - 1); notifyDataSetChanged(); } @Override public int getItemViewType(int position) { return mSeparatorsSet.contains(position) ? TYPE_SEPARATOR : TYPE_ITEM; } holder = new ViewHolder(); switch (type) { case TYPE_ITEM: emailView= inflater.inflate(R.layout.email_listview_row, null); break; case TYPE_SEPARATOR: emailView= inflater.inflate(R.layout.item2, null); holder.textView = (TextView)emailView.findViewById(R.id.textSeparator); emailView.setTag(holder); holder.textView.setText("SEPARATOR"); break; } Here is ViewHolder class public static class ViewHolder { public TextView textView; } if anybody knows then please tell me where I am doing wrong. Thanx

    Read the article

  • Restart list numbering in word for each new list created

    - by Feena
    Hi, I am exporting content from a jsp page into MS Word using javascript. When the user is in Word there is a table with 10 rows and 2 columns, A & B. The user creates an ordered list in row 1, column A like this: 1 dog 2 cat 3 mouse if the user then creates a second list in row 1 column B is turns out like this: 4 car 5 truck 6 bike instead of: 1 car 2 truck 3 bike Word is set up to continue the numbering in lists from prior lists automatically. I know this can be reset easily but the users dont want to have to do this. They want the numbering of any potential lists created to restarted at 1. when the document is exported into Word and opened in front of them. So this must be set up in the javasctipt code or using a style or something prior to getting into Word. This is what I dont know how to do. Any help is much appreciated. Thanks, Feena.

    Read the article

  • Django: getting the list of related records for a list of objects

    - by Silver Light
    Hello! I have two models related one-to many: class Person(models.Model): name = models.CharField(max_length=255); surname = models.CharField(max_length=255); age = models.IntegerField(); class Dog(models.Model): name = models.CharField(max_length=255); owner = models.ForeignKey('Person'); I want to output a list of persons below each person a list of dogs he has. Here's how I can do it: in view: persons = Person.objects.all()[0:100]; in template: {% for p in persons %} {{ p.name }} has dogs:<br /> {% for d in persons.dog_set.all %} - {{ d.name }}<br /> {% endfor %} {% endfor %} But if I do it like that, Django will execute 101 SQL queries which is very inefficient. I tried to make a custom manager, which will get all the persons, then all the dogs and links them in python, but then I can't use paginator (my another question: http://stackoverflow.com/questions/2532475/django-paginator-raw-sql-query ) and it looks quite ugly. Is there a more graceful way doing this?

    Read the article

  • Trying to get these list items to display inline

    - by Joel
    I have several unordered lists that I want to display like this: <ul> <li><img></li> <li><p></li> //inline </ul> //linebreak <ul> <li><img></li> <li><p></li> //inline </ul> ...etc I'm not able to get the li items to be inline with eachother. They are stacking vertically. I have stripped away most styling but still can't figure out what I'm doing wrong: html: <ul class="instrument"> <li class="imagebox"><img src="/images/matepe.jpg" width="247" height="228" alt="Matepe" /></li> <li class="textbox"><p>The matepe is a 24 key instrument that is played by the Kore-Kore people in North-Eastern Zimbabwe and Mozambique. It utilizes four fingers-each playing an individual melody. These melodies also interwieve to create resultant melodies that can be manipulated thru accenting different fingers. The matepe is used in Rattletree as the bridge from the physical world to the spirit world. The matepe is used in the Kore-Kore culture to summon the Mhondoro spirits which are thought to be able to communicate directly with Mwari (God) without the need of an intermediary.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/soprano_little.jpg" border="0" width="247" height="170" alt="Soprano" /></li> <li class="textbox"><p>The highest voice of the Rattletree Marimba orchestra is the Soprano marimba. The soprano is used to whip up the energy on the dancefloor and help people reach ecstatic state with it's high and clear singing voice. The range of these sopranos goes much lower than 'typical' Zimbabwean style sopranos. The sopranos play the range of the right hand of the matepe and go two notes higher and five notes lower. Rattletree uses two sopranos.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/bari_little.jpg" border="0" width="247" height="170" alt="Baritone" /></li> <li class="textbox"><p>The Baritone is the next lower voice in the orchestra. The bari is where the funk is. Generally bubbling over the Bass line, the baritone creates the syncopations and polyrhythms that messes with the 'minds' of the dancers and helps seperate the listener from the physical realm of thought. The range of the baritone covers the full range of the left hand side of the matepe.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/darren_littlebass.jpg" border="0" width="247" height="195" alt="Bass"/><strong>Bass Marimba</strong></li> <li class="textbox"><p>The towering Bass Marimba is the foundation of the Rattletree Marimba sound. Putting out frequencies as low as 22hZ, the bass creates the drive that gets the dancefloor moving. It is 5.5' tall, 9' long, and 4' deep. It is played by standing on a platform and struck with mallets that have lacross-ball size heads (they are actually made with rubber dog balls). The Bass marimba's range covers the lowest five notes of the matepe and goes another five notes lower.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/wayne_little.jpg" border="0" width="247" height="177" alt="Drums"/><strong>Drumset</strong></li> <li class="textbox"><p>All the intricate polyrhythms are held together tastefully with the drumset. The drums provides the consistancy and grounding that the dancers need to keep going all night. While the steady kick and high-hat provide that grounding function, the toms and snare and allowed to be another voice in the poylrhythmic texture-helping the dancers abandon the concept of a "one" within this cyclical music.</p></li> </ul> css: ul.instrument { text-align:left; display:inline; } ul.instrument li { list-style-type: none; } li.imagebox { } li.textbox { } li.textbox p{ width: 247px; }

    Read the article

  • Link List Implementation Help - Visual C++

    - by Greenhouse Gases
    Hi there I'm trying to implement a link list which stores the city name (though you will see this commented out as I need to resolve the issue of not being able to use string and needing to use a primitive data type instead during the declaration), longitude, latitude and of course a pointer to the next node in the chain. I am new to the Visual C++ environment and my brain is somewhat scrambled after coding for several straight hours today so I wondered if anyone could help resolve the 2 errors I am getting (ignore the #include syntax as I had to change them to avoid the browser interpreting html!): 1U08221.obj : error LNK2028: unresolved token (0A000298) "public: __thiscall Locations::Locations(void)" (??0Locations@@$$FQAE@XZ) referenced in function "int __clrcall main(cli::array^)" (?main@@$$HYMHP$01AP$AAVString@System@@@Z) 1U08221.obj : error LNK2019: unresolved external symbol "public: __thiscall Locations::Locations(void)" (??0Locations@@$$FQAE@XZ) referenced in function "int __clrcall main(cli::array^)" (?main@@$$HYMHP$01AP$AAVString@System@@@Z) The code for my header file is here: include string struct locationNode { //char[10] nodeCityName; double nodeLati; double nodeLongi; locationNode* Next; }; class Locations { private: int size; public: Locations(); // constructor for the class locationNode* Head; int Add(locationNode* Item); }; and here is the code for the file containing the main method: // U08221.cpp : main project file. include "stdafx.h" include "Locations.h" include iostream include string using namespace std; int n = 0; int x; string cityNameInput; bool acceptedInput = false; int Locations::Add(locationNode *NewItem) { locationNode *Sample = new locationNode; Sample = NewItem; Sample-Next = Head; Head = Sample; return size++; } void CorrectCase(string name) // Correct upper and lower case letters of input { x = name.size(); int firstLetVal = name[0], letVal; n = 1; // variable for name index from second letter onwards if((name[0] 90) && (name[0] < 123)) // First letter is lower case { firstLetVal = firstLetVal - 32; // Capitalise first letter name[0] = firstLetVal; } while(n <= x - 1) { if((name[n] = 65) && (name[n] <= 90)) { letVal = name[n] + 32; name[n] = letVal; } n++; } cityNameInput = name; } void nameValidation(string name) { n = 0; // start from first letter x = name.size(); while(!acceptedInput) { if((name[n] = 65) && (name[n] <= 122)) // is in the range of letters { while(n <= x - 1) { while((name[n] =91) && (name[n] <=97)) // ERROR!! { cout << "Please enter a valid city name" << endl; cin name; } n++; } } else { cout << "Please enter a valid city name" << endl; cin name; } if(n <= x - 1) { acceptedInput = true; } } cityNameInput = name; } int main(array ^args) { cout << "Enter a city name" << endl; cin cityNameInput; nameValidation(cityNameInput); // check is made up of valid characters CorrectCase(cityNameInput); // corrects name to standard format of capitalised first letter, and lower case subsequent letters cout << cityNameInput; cin cityNameInput; Locations::Locations(); Locations *Parts = new Locations(); locationNode *Part; Part = new locationNode; //Part-nodeCityName = "London"; Part-nodeLati = 87; Part-nodeLongi = 80; Parts-Add(Part); } I am familiar with the concepts but somewhat inexperienced with OOP so am making some silly errors that you can never find when you've stared at something too long. Any help you can offer will be appreciated! Thanks

    Read the article

  • list images from directory by function

    - by osc2nuke
    i'm using this function: function getmyimages($qid){ $imgdir = 'modules/Projects/uploaded_project_images/'. $qid .''; // the directory, where your images are stored $allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show $dimg = opendir($imgdir); while($imgfile = readdir($dimg)) { if(in_array(strtolower(substr($imgfile,-3)),$allowed_types)) { $a_img[] = $imgfile; sort($a_img); reset ($a_img); } } $totimg = count($a_img); // total image number for($x=0; $x < $totimg; $x++) { $size = getimagesize($imgdir.'/'.$a_img[$x]); // do whatever $halfwidth = ceil($size[0]/2); $halfheight = ceil($size[1]/2); $mytest = 'name: '.$a_img[$x].' width: '.$size[0].' height: '.$size[1].'<br /><a href="'. $imgdir .'/'.$a_img[$x].'">'. $a_img[$x]. '</a>'; } return $mytest; } And i call this function between a while row as: $sql_select = $db->sql_query('SELECT * from '.$prefix.'_projects WHERE topic=\''.$cid.'\''); OpenTable(); while ($row2 = $db->sql_fetchrow($sql_select)){ $qid = $row2['qid']; $project_query = $db->sql_query('SELECT p.uid, p.uname, p.subject, p.story, p.storyext, p.date, p.topic, p.pdate, p.materials, p.bidoptions, p.projectduration, pd.id_duration, pm.material_id, pbo.bidid, pc.cid FROM ' . $prefix . '_projects p, ' . $prefix . '_projects_duration pd, ' . $prefix . '_project_materials pm, ' . $prefix . '_project_bid_options pbo, ' . $prefix . '_project_categories pc WHERE p.topic=\''.$cid.'\' and p.qid=\''.$qid.'\' and p.bidoptions=pbo.bidid and p.materials=pm.material_id and p.projectduration=pd.id_duration'); while ($project_row = $db->sql_fetchrow($project_query)) { //$qid = $project_row['qid']; $uid = $project_row['uid']; $uname = $project_row['uname']; $subject = $project_row['subject']; $story = $project_row['story']; $storyext = $project_row['storyext']; $date = $project_row['date']; $topic = $project_row['topic']; $pdate = $project_row['pdate']; $materials = $project_row['materials']; $bidoptions = $project_row['bidoptions']; $projectduration = $project_row['projectduration']; //Get the topic name $topic_query = $db->sql_query('SELECT cid,title from '.$prefix.'_project_categories WHERE cid =\''.$cid.'\''); while ($topic_row = $db->sql_fetchrow($topic_query)) { $topic_id = $topic_row['cid']; $topic_title = $topic_row['title']; } //Get the material text $material_query = $db->sql_query('SELECT material_id,material_name from '.$prefix.'_project_materials WHERE material_id =\''.$materials.'\''); while ($material_row = $db->sql_fetchrow($material_query)) { $material_id = $material_row['material_id']; $material_name = $material_row['material_name']; } //Get the bid methode $bid_query = $db->sql_query('SELECT bidid,bidname from '.$prefix.'_project_bid_options WHERE bidid =\''.$bidoptions.'\''); while ($bid_row = $db->sql_fetchrow($bid_query)) { $bidid = $bid_row['bidid']; $bidname = $bid_row['bidname']; } //Get the project duration $duration_query = $db->sql_query('SELECT id_duration,duration_value,duration_alias from '.$prefix.'_projects_duration WHERE id_duration =\''.$projectduration.'\''); while ($duration_row = $db->sql_fetchrow($duration_query)) { $id_duration = $duration_row['id_duration']; $duration_value = $duration_row['duration_value']; $duration_alias = $duration_row['duration_alias']; } } echo '<br/><b>id</b>--->' .$qid. '<br/><b>uid</b>--->' .$uid. '<br/><b>username</b>--->' .$uname. '<br/><b>subject</b>--->'.$subject. '<br/><b>story1</b>--->'.$story. '<br/><b>story2</b>--->'.$storyext. '<br/><b>postdate</b>--->'.$date. '<br/><b>categorie</b>--->'.$topic_title . '<br/><b>project start</b>--->'.$pdate. '<br/><b>materials</b>--->'.$material_name. '<br/><b>bid methode</b>--->'.$bidname. '<br/><b>project duration</b>--->'.$duration_alias.'<br /><br /><br/><b>image url</b>--->'.getmyimages($qid).'<br /><br />'; } CloseTable(); the result outputs only the "last" file from the directories. if i do a echo instead of a return $mytest; it read the whole directory but ruïns the output.

    Read the article

  • SPSiteDataQuery Returns Only One List Type At A Time

    - by Brian Jackett
    The SPSiteDataQuery class in SharePoint 2007 is very powerful, but it has a few limitations.  One of these limitations that I ran into this morning (and caused hours of frustration) is that you can only return results from one list type at a time.  For example, if you are trying to query items from an out of the box custom list (list type = 100) and document library (list type = 101) you will only get items from the custom list (SPSiteDataQuery defaults to list type = 100.)  In my situation I was attempting to query multiple lists (created from custom list templates 10001 and 10002) each with their own content types. Solution     Since I am only able to return results from one list type at a time, I was forced to run my query twice with each time setting the ServerTemplate (translates to ListTemplateId if you are defining custom list templates) before executing the query.  Below is a snippet of the code to accomplish this. SPSiteDataQuery spDataQuery = new SPSiteDataQuery(); spDataQuery.Lists = "<Lists ServerTemplate='10001' />"; // ... set rest of properties for spDataQuery   var results = SPContext.Current.Web.GetSiteData(spDataQuery).AsEnumerable();   // only change to SPSiteDataQuery is Lists property for ServerTemplate attribute spDataQuery.Lists = "<Lists ServerTemplate='10002' />";   // re-execute query and concatenate results to existing entity results = results.Concat(SPContext.Current.Web.GetSiteData(spDataQuery).AsEnumerable());   Conclusion     Overall this isn’t an elegant solution, but it’s a workaround for a limitation with the SPSiteDataQuery.  I am now able to return data from multiple lists spread across various list templates.  I’d like to thank those who commented on this MSDN page that finally pointed out the limitation to me.  Also a thanks out to Mark Rackley for “name dropping” me in his latest article (which I humbly insist I don’t belong in such company)  as well as encouraging me to write up a quick post on this issue above despite my busy schedule.  Hopefully this post saves some of you from the frustrations I experienced this morning using the SPSiteDataQuery.  Until next time, Happy SharePoint’ing all.         -Frog Out   Links MSDN Article for SPSiteDataQuery http://msdn.microsoft.com/en-us/library/microsoft.sharepoint.spsitedataquery.lists.aspx

    Read the article

  • 75 Top Open Source Security Apps

    <b>Datamation:</b> "This year, we've once again updated our list of top open source security apps. While the list isn't exhaustive by any means, we tried to include many of the best tools in a variety of categories."

    Read the article

  • Top 31 Favorite Features in Windows Server 2012

    - by KeithMayer
    Over the past month, my fellow IT Pro Technical Evangelists and I have authored a series of articles about our Top 31 Favorite Features in Windows Server 2012.  Now that our series is complete, I’m providing a clickable index below of all of the articles in the series for your convenience, just in case you perhaps missed any of them when they were first released.  Hope you enjoy our Favorite Features in Windows Server 2012! Top 31 Favorite Features in Windows Server 2012 The Cloud OS Platform by Kevin Remde Server Manager in Windows Server 2012 by Brian Lewis Feel the Power of PowerShell 3.0 by Matt Hester Live Migrate Your VMS in One Line of PowerShell by Keith Mayer Windows Server 2012 and Hyper-V Replica by Kevin Remde Right-size IT Budgets with “Storage Spaces” by Keith Mayer Yes, there is an “I” in Team – the NIC Team! by Kevin Remde Hyper-V Network Virtualization by Keith Mayer Get Happy over the FREE Hyper-V Server 2012 by Matt Hester Simplified BranchCache in Windows Server 2012 by Brian Lewis Getting Snippy with PowerShell 3.0 by Matt Hester How to Get Unbelievable Data Deduplication Results by Chris Henley of Veeam Simplified VDI Configuration and Management by Brian Lewis Taming the New Task Manager by Keith Mayer Improve File Server Resiliency with ReFS by Keith Mayer Simplified DirectAccess by Sumeeth Evans SMB 3.0 – The Glue in Windows Server 2012 by Matt Hester Continuously Available File Shares by Steven Murawski of Edgenet Server Core - Improved Taste, Less Filling, More Uptime by Keith Mayer Extend Your Hyper-V Virtual Switch by Kevin Remde To NIC or to Not NIC Hardware Requirements by Brian Lewis Simplified Licensing and Server Versions by Kevin Remde I Think, Therefore IPAM! by Kevin Remde Windows Server 2012 and the RSATs by Kevin Remde Top 3 New Tricks in the Active Directory Admin Center by Keith Mayer Dynamic Access Control by Brian Lewis Get the Gremlin out of Your Active Directory Virtualized Infrastructure by Matt Hester Scoping out the New DHCP Failover by Keith Mayer Gone in 8 Seconds – The New CHKDSK by Matt Hester New Remote Desktop Services (RDS) by Brian Lewis No Better Time Than Now to Choose Hyper-V by Matt Hester What’s Next? Keep Learning! Want to learn more about Windows Server 2012 and Hyper-V Server 2012?  Want to prepare for certification on Windows Server 2012? Do It: Join our Windows Server 2012 “Early Experts” Challenge online peer study group for FREE at http://earlyexperts.net. You’ll get FREE access to video-based lectures, structured study materials and hands-on lab activities to help you study and prepare!  Along the way, you’ll be part of an IT Pro community of over 1,000+ IT Pros that are all helping each other learn Windows Server 2012! What are Your Favorite Features? Do you have a Favorite Feature in Windows Server 2012 that we missed in our list above?  Feel free to share your favorites in the comments below! Keith Build Your Lab! Download Windows Server 2012 Don’t Have a Lab? Build Your Lab in the Cloud with Windows Azure Virtual Machines Want to Get Certified? Join our Windows Server 2012 "Early Experts" Study Group

    Read the article

< Previous Page | 27 28 29 30 31 32 33 34 35 36 37 38  | Next Page >