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  • Intel IA-32 Assembly

    - by Kay
    I'm having a bit of difficulty converting the following java code into Intel IA-32 Assembly: class Person() { char name [8]; int age; void printName() {...} static void printAdults(Person [] list) { for(int k = 0; k < 100; k++){ if (list[k].age >= 18) { list[k].printName(); } } } } My attempt is: Person: push ebp; save callers ebp mov ebp, esp; setup new ebp push esi; esi will hold name push ebx; ebx will hold list push ecx; ecx will hold k init: mov esi, [ebp + 8]; mov ebx, [ebp + 12]; mov ecx, 0; k=0 forloop: cmp ecx, 100; jge end; if k>= 100 then break forloop cmp [ebx + 4 * ecx], 18 ; jl auxloop; if list[k].age < 18 then go to auxloop jmp printName; printName: auxloop: inc ecx; jmp forloop; end: pop ecx; pop ebx; pop esi; pop ebp; Is my code correct? NOTE: I'm not allowed to use global variables.

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  • A shortest path problem with superheroes and intergalactic journeys

    - by Dman
    You are a super-hero in the year 2222 and you are faced with this great challenge: starting from your home planet Ilop you must try to reach Acinhet or else your planet will be destroyed by evil green little monsters. To do this you are given a map of the universe: there are N planets and M inter-planetary connections ( bidirectional ) that bind these planets. Each connection requires a certain time and a certain amount of fuel in order for you to cover the connection from one planet to another. The total time spent going from one planet to another is obtained by multiplying the time past to cover each connection between all the planets you go through. There are some "key planets", that allow you to refuel if you arrive on those certain "key planets". A "key planet" is the planet with the property that if it disappears the road between at least two planets would be lost.(In the example posted below with the input/output files such a "key planet" is 2 because without it the road to 7 would be lost) When you start your mission you are given the possibility of choosing between K ships each with its own maximum fuel capacity. The goal is to find the SHORTEST TIME CONSUMING path but also choose the ship with the minimum fuel capacity that can cover that shortest path(this means that if more ships can cover the shortest path you choose the one with the minimum fuel capacity). Because the minimum time can be a rather large number (over long long int) you are asked to provide only the last 6 digits of the number. For a better understanding of the task, here is an example of input/output files: INPUT: mission.in 7 8 6 1 4 6 5 9 8 7 10 1 2 7 8 1 4 14 9 1 5 3 1 2 3 1 2 2 7 7 1 3 4 2 2 4 6 4 1 5 6 3 7 On the first line (in order): N M K On the second line :the number for the starting planet and the finishing planet On the third line :K numbers that represent the capacities of the ships you can choose from Then you have M lines, all of them have the same structure: Xi Yi Ti Fi-which means that there is a connection between Xi and Yi and you can cover the distance from Xi to Yi in Ti time and with a Fi fuel consumption. OUTPUT:mission.out 000014 8 1 2 3 4 On the first line:the minimum time and fuel consumption; On the second line :the path Restrictions: 2 = N = 1 000 1 = M = 30 000 1 = K = 10 000 Any suggestions or ideas of how this problem might be solved would be most welcomed.

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  • JOptionPane opening another JFrame

    - by mike_hornbeck
    So I'm continuing my fight with this : http://stackoverflow.com/questions/2923545/creating-java-dialogs/2926126 task. Now my JOptionPane opens new window with envelope overfiew, but I can't change size of this window. Also I wanted to have sender's data in upper left corner, and receiver's data in bottom right. How can I achieve that ? There is also issue with OptionPane itself. After I click 'OK' it opens small window in the upper left corner of the screen. What is this and why it's appearing ? My code: import java.awt.*; import java.awt.Font; import javax.swing.*; public class Main extends JFrame { private static JTextField nameField = new JTextField(20); private static JTextField surnameField = new JTextField(); private static JTextField addr1Field = new JTextField(); private static JTextField addr2Field = new JTextField(); private static JComboBox sizes = new JComboBox(new String[] { "small", "medium", "large", "extra-large" }); public Main(){ JPanel mainPanel = new JPanel(); mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); getContentPane().add(mainPanel); JPanel addrPanel = new JPanel(new GridLayout(0, 1)); addrPanel.setBorder(BorderFactory.createTitledBorder("Receiver")); addrPanel.add(new JLabel("Name")); addrPanel.add(nameField); addrPanel.add(new JLabel("Surname")); addrPanel.add(surnameField); addrPanel.add(new JLabel("Address 1")); addrPanel.add(addr1Field); addrPanel.add(new JLabel("Address 2")); addrPanel.add(addr2Field); mainPanel.add(addrPanel); mainPanel.add(new JLabel(" ")); mainPanel.add(sizes); String[] buttons = { "OK", "Cancel"}; int c = JOptionPane.showOptionDialog( null, mainPanel, "My Panel", JOptionPane.DEFAULT_OPTION, JOptionPane.PLAIN_MESSAGE, null, buttons, buttons[0] ); if(c ==0){ new Envelope(nameField.getText(), surnameField.getText(), addr1Field.getText() , addr2Field.getText(), sizes.getSelectedIndex()); } setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); pack(); setVisible(true); } public static void main(String[] args) { new Main(); } } class Envelope extends JFrame { private final int SMALL=0; private final int MEDIUM=1; private final int LARGE=2; private final int XLARGE=3; public Envelope(String n, String s, String a1, String a2, int i){ Container content = getContentPane(); JPanel mainPanel = new JPanel(); mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); mainPanel.add(new JLabel("John Doe")); mainPanel.add(new JLabel("FooBar str 14")); mainPanel.add(new JLabel("Newark, 45-99")); JPanel dataPanel = new JPanel(); dataPanel.setFont(new Font("sansserif", Font.PLAIN, 32)); //set size from i mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); mainPanel.setBackground(Color.ORANGE); mainPanel.add(new JLabel("Mr "+n+" "+s)); mainPanel.add(new JLabel(a1)); mainPanel.add(new JLabel(a2)); content.setSize(450, 600); content.setBackground(Color.ORANGE); content.add(mainPanel, BorderLayout.NORTH); content.add(dataPanel, BorderLayout.SOUTH); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); pack(); setVisible(true); } }

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  • How to write a function to output unconstant loop

    - by tunpishuang
    Here is the function description test($argv) $argv is an array, for example $argv=array($from1,$to1,$from2,$to2.....); array items must be even. $argv=array(1,2,4,5) : this will output values like below: 1_4 1_5 2_4 2_5 The number of array $argv's is not constant. Maybe 3 or 4 levels of loop will be outputed. I know this will used RECURSIVE , but i don't know exactly how to code.

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  • javascript robot

    - by sarah
    hey guys! I need help making this robot game in javascript (notepad++) please HELP! I'm really confused by the functions <html> <head><title>Robot Invasion 2199</title></head> <body style="text-align:center" onload="newGame();"> <h2>Robot Invasion 2199</h2> <div style="text-align:center; background:white; margin-right: auto; margin-left:auto;"> <div style=""> <div style="width: auto; border:solid thin red; text-align:center; margin:10px auto 10px auto; padding:1ex 0ex;font-family: monospace" id="scene"></pre> </div> <div><span id="status"></span></div> <form style="text-align:center"> PUT THE CONTROL PANEL HERE!!! </form> </div> <script type="text/javascript"> // GENERAL SUGGESTIONS ABOUT WRITING THIS PROGRAM: // You should test your program before you've finished writing all of the // functions. The newGame, startLevel, and update functions should be your // first priority since they're all involved in displaying the initial state // of the game board. // // Next, work on putting together the control panel for the game so that you // can begin to interact with it. Your next goal should be to get the move // function working so that everything else can be testable. Note that all nine // of the movement buttons (including the pass button) should call the move // function when they are clicked, just with different parameters. // // All the remaining functions can be completed in pretty much any order, and // you'll see the game gradually improve as you write the functions. // // Just remember to keep your cool when writing this program. There are a // bunch of functions to write, but as long as you stay focused on the function // you're writing, each individual part is not that hard. // These variables specify the number of rows and columns in the game board. // Use these variables instead of hard coding the number of rows and columns // in your loops, etc. // i.e. Write: // for(i = 0; i < NUM_ROWS; i++) ... // not: // for(i = 0; i < 15; i++) ... var NUM_ROWS = 15; var NUM_COLS = 25; // Scene is arguably the most important variable in this whole program. It // should be set up as a two-dimensional array (with NUM_ROWS rows and // NUM_COLS columns). This represents the game board, with the scene[i][j] // representing what's in row i, column j. In particular, the entries should // be: // // "." for empty space // "R" for a robot // "S" for a scrap pile // "H" for the hero var scene; // These variables represent the row and column of the hero's location, // respectively. These are more of a conveniece so you don't have to search // for the "H" in the scene array when you need to know where the hero is. var heroRow; var heroCol; // These variables keep track of various aspects of the gameplay. // score is just the number of robots destroyed. // screwdrivers is the number of sonic screwdriver charges left. // fastTeleports is the number of fast teleports remaining. // level is the current level number. // Be sure to reset all of these when a new game starts, and update them at the // appropriate times. var score; var screwdrivers; var fastTeleports; var level; // This function should use a sonic screwdriver if there are still charges // left. The sonic screwdriver turns any robot that is in one of the eight // squares immediately adjacent to the hero into scrap. If there are no charges // left, then this function should instead pop up a dialog box with the message // "Out of sonic screwdrivers!". As with any function that alters the game's // state, this function should call the update function when it has finished. // // Your "Sonic Screwdriver" button should call this function directly. function screwdriver() { // WRITE THIS FUNCTION } // This function should move the hero to a randomly selected location if there // are still fast teleports left. This function MUST NOT move the hero on to // a square that is already occupied by a robot or a scrap pile, although it // can move the hero next to a robot. The number of fast teleports should also // be decreased by one. If there are no fast teleports left, this function // should just pop up a message box saying so. As with any function that alters // the game's state, this function should call the update function when it has // finished. // // HINT: Have a loop that keeps trying random spots until a valid one is found. // HINT: Use the validPosition function to tell if a spot is valid // // Your "Fast Teleport" button s

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  • javascript too much recursion?

    - by Ken
    Hi, I'm trying to make a script that automatically starts uploading after the data has been enter in the database(I need the autoId that the database makes to upload the file). When I run the javascript the scripts runs the php file but it fails calling the other php to upload the file. too much recursion setTimeout(testIfToegevoegd(),500); the script that gives the error send("/projects/backend/nieuwDeeltaak.php",'deeltaakNaam='+f.deeltaaknaam.value+'&beschrijving='+ f.beschrijving.value+'&startDatum='+f.startDatum.value+'&eindDatum='+f.eindDatum.value +'&deeltaakLeider='+f.leiderID.value+'&projectID='+f.projectID.value,id); function testIfToegevoegd(){ if(document.getElementById('resultaat').innerHTML == "<b>De deeltaak werd toegevoegd</b>"){ //stop met testen + upload file document.getElementById('nieuwDeeltaak').target = 'upload_target'; document.forms["nieuwDeeltaak"].submit() }else{ setTimeout(testIfToegevoegd(),500); } } testIfToegevoegd(); sorry for the dutch names we have to use them it is a school project. when I click the button that calls all this for a second time (after the error) it works fine.

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  • Need Ontology of software engineering terms

    - by v.rekha
    I'm trying to implement a Question Answering System based on software engineering ontology. This is a [class/university?] project; it will use the java language. Can you please help me locate software engineering ontologies i.e. ontologies [or even taxonomies / folksonomies ?] that include words and concepts found in the domain of Software Engineering. Editor's note: I took the liberty of rewriting this question. It was initially poorly worded and being misunderstood was closed. Maybe it can be reopen, for it is programming related and of interest to some SO contributors. According to this dupe: the poster is looking for an OWL resource: quote, "I need software engineering ontology.owl file."

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  • Prolog - generate correct bracketing

    - by Henrik Bak
    I'd like to get some help in the following exam problem, i have no idea how to do this: Input: a list of numbers, eg.: [1,2,3,4] Output: every possible correct bracketing. Eg.: (in case of input [1,2,3,4]): ((1 2) (3 4)) ((1 (2 3)) 4) (1 ((2 3) 4)) (1 (2 (3 4))) (((1 2) 3) 4) Bracketing here is like a method with two arguments, for example multiplication - then the output is the possible multiplication orders. Please help, i'm stuck with this one. Any help is appreciated, thanks!

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  • total number of magic square from 9 numbers

    - by Peeyush
    9 numbers need to be arranged in a magic number square. A magic number square is a square of numbers that is arranged such that every row and column has the same sum.(condition for diagonal has been relaxed) For example: 1 2 3 3 2 1 2 2 2 How do we calculate total number of distinct magic square from 9 numbers. Two magic number squares are distinct if they differ in value at one or more positions. For example, there is only one magic number square that can be made of 9 instances of the same number. e.g. for these 9 numbers { 4, 4, 4, 4, 4, 4, 4, 4, 4 }, answer should be 1. Also the complexity should be optimal. Do we need to iterate through all the permutations , discarding if a[0]+a[1]+a[2] %3!=0 such combinations ? moreover how do we remove duplicate magic square?

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  • How do I check the validity of the Canadian Social Insurance Number in C#?

    - by user518307
    I've been given the assignment to write an algorithm in C# that checks the validity of a Canadian Social Insurance Number (SIN). Here are the steps to validate a SIN. Given an example Number: 123 456 782 Remove the check digit (the last digit): 123456782 Extract the even digits (2,4,6,8th digith): 12345678 Double them: 2 4 6 8 | | | | v v v v 4 8 12 16 Add the digits together: 4+8+1+2+1+6 = 22 Add the Odd placed digits: 1+3+5+7 = 16 Total : 38 Validity Algorithm If the total is a multiple of 10, the check digit should be zero. Otherwise, Subtract the Total from the next highest multiple of 10 (40 in this case) The check digit for this SIN must be equal to the difference of the number and the totals from earlier (in this case, 40-38 = 2; check digit is 2, so the number is valid) I'm lost on how to actually implement this in C#, how do I do this?

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  • class header+ implementation

    - by igor
    what am I doing wrong here? I keep on getting a compilation error when I try to run this in codelab (turings craft) Instructions: Write the implementation (.cpp file) of the GasTank class of the previous exercise. The full specification of the class is: A data member named amount of type double. A constructor that no parameters. The constructor initializes the data member amount to 0. A function named addGas that accepts a parameter of type double . The value of the amount instance variable is increased by the value of the parameter. A function named useGas that accepts a parameter of type double . The value of the amount data member is decreased by the value of the parameter. A function named getGasLevel that accepts no parameters. getGasLevel returns the value of the amount data member. class GasTank{ double amount; GasTank(); void addGas(double); void useGas(double); double getGasLevel();}; GasTank::GasTank(){ amount=0;} double GasTank::addGas(double a){ amount+=a;} double GasTank::useGas(double a){ amount+=a;} double GasTank::getGasLevel(){ return amount;}

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  • A[i] * A[j] = k in O(nlog(n))

    - by gleb-pendler
    A is an Array of n positive int numbers k given int Algorithm should find if there is a pair of numbers which product gives the result a. A[i] * A[j] = k b. A[i] = A[j] + k if there is such a couple the algorithm should return thier index. thanks in advance.

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  • Tricky Big-O complexity

    - by timeNomad
    public void foo (int n, int m) { int i = m; while (i > 100) i = i/3; for (int k=i ; k>=0; k--) { for (int j=1; j<n; j*=2) System.out.print(k + "\t" + j); System.out.println(); } } I figured the complexity would be O(logn). That is as a product of the inner loop, the outer loop -- will never be executed more than 100 times, so it can be omitted. What I'm not sure about is the while clause, should it be incorporated into the Big-O complexity? For very large i values it could make an impact, or arithmetic operations, doesn't matter on what scale, count as basic operations and can be omitted?

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  • How should I generate the partitions / pairs for the Chinese Postman problem?

    - by Simucal
    I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own. The part that is giving me trouble is generating the partitions of pairings for the odd vertices. For example, if I had the following labeled odd verticies in a graph: 1 2 3 4 5 6 I need to find all the possible pairings / partitions I can make with these vertices. I've figured out I'll have i paritions given: n = num of odd verticies k = n / 2 i = ((2k)(2k-1)(2k-2)...(k+1))/2 So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions. The 15 partions would look like: 1 2 3 4 5 6 1 2 3 5 4 6 1 2 3 6 4 5 ... 1 6 ... Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition). These represent the edges the postman will have to walk twice. At first I thought I had worked out an appropriate algorithm for generating these partitions / pairs but it is flawed. I found it wasn't a simple permutation/combination problem. Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?

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  • Java - Creating a Compiler Help

    - by Brian
    So for my programming class we have had a project to create a virtual machine including a memory unit, cpu, Input, Output, Instruction Register, Program Counter, MAR, MDR and so on. Now we need to create a compiler using Java Code that will take a .exe file written in some txt editor and convert it to java byte code and run the code. The code we will be writing in the .exe file is machine code along the lines of: IN X IN Y ADD X STO Y OUT Y STOP DC X 0 DC Y 0 I am just a beginner and only have 2 days to write this and am very lost and have no idea where to start....Any Help will be much appreciated. Thanks

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Why is my file being cleared if I don't save it?

    - by Kat
    My program is suppose to maintain a collection of Photos in a PhotoAlbum. It begins by reading a folder of photos and adds them to my PhotoAlbum. It then prints a menu that allows the user to list all the photos, add a photo, find a photo, save, and quit the program. Right now if I run my program it will add the 100 photos to the PhotoAlbum, but if I quit the program without saving, it clears the file I am reading from even if I haven't added a photo or done anything to the PhotoAlbum and I'm not sure why. Here is my method for printing to a file: private static void saveFile(PrintWriter writer) { String result; ArrayList<Photo> temp = album.getPhotoAlbum(); for (int i = 0; i < temp.size(); i++){ result = temp.get(i).toString() + "\n"; writer.println(result); } writer.close(); } And where the PrintWriter is instantiated: File file = new File(args[0] + File.separator + "album.dat"); try { PrintWriter fout = new PrintWriter(new FileWriter(file)); fileWriter = fout; } catch (IOException e){ System.out.println("ReadFromFile: Folder " + args[0] + " is not found."); System.exit(0); } And where it is called in my runMenu Method: private static void runMainMenu(Scanner scan) { String input; do { showMainMenu(); input = scan.nextLine().toLowerCase(); switch (input.charAt(0)) { case 'p': System.out.println(album.toString()); break; case 'a': album.addPhoto(readPhoto(scan, t)); break; case 'f': findMenu(scan); break; case 's': saveFile(fileWriter); System.exit(0); break; case 'q': break; default: System.out.println("Invalid entry: " + input.charAt(0)); break; } } while (!input.equalsIgnoreCase("q")); }

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  • Validating an integer or String without try-catch

    - by Phil
    Ok, I'm lost. I am required to figure out how to validate an integer and String, but for some stupid reason, I can't use the Try-Catch method. I know this is the easiest way and so all the solutions on the internet are using it. I'm writing in Java. The deal is this, I need someone to put in an numerical ID and String name. If either one of the two inputs are invalid I must tell them they made a mistake. Can someone help me?

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  • Generalizing Fibonacci sequence with SICStus Prolog

    - by Christophe Herreman
    I'm trying to find a solution for a query on a generalized Fibonacci sequence (GFS). The query is: are there any GFS that have 885 as their 12th number? The initial 2 numbers may be restricted between 1 and 10. I already found the solution to find the Nth number in a sequence that starts at (1, 1) in which I explicitly define the initial numbers. Here is what I have for this: fib(1, 1). fib(2, 1). fib(N, X) :- N #> 1, Nmin1 #= N - 1, Nmin2 #= N - 2, fib(Nmin1, Xmin1), fib(Nmin2, Xmin2), X #= Xmin1 + Xmin2. For the query mentioned I thought the following would do the trick, in which I reuse the fib method without defining the initial numbers explicitly since this now needs to be done dynamically: fib2 :- X1 in 1..10, X2 in 1..10, fib(1, X1), fib(2, X2), fib(12, 885). ... but this does not seem to work. Is it not possible this way to define the initial numbers, or am I doing something terribly wrong? I'm not asking for the solution, but any advice that could help me solve this would be greatly appreciated.

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  • C++ Operator overloading - 'recreating the Vector'

    - by Wallter
    I am currently in a collage second level programing course... We are working on operator overloading... to do this we are to rebuild the vector class... I was building the class and found that most of it is based on the [] operator. When I was trying to implement the + operator I run into a weird error that my professor has not seen before (apparently since the class switched IDE's from MinGW to VS express...) (I am using Visual Studio Express 2008 C++ edition...) Vector.h #include <string> #include <iostream> using namespace std; #ifndef _VECTOR_H #define _VECTOR_H const int DEFAULT_VECTOR_SIZE = 5; class Vector { private: int * data; int size; int comp; public: inline Vector (int Comp = 5,int Size = 0) : comp(Comp), size(Size) { if (comp > 0) { data = new int [comp]; } else { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; } } int size_ () const { return size; } int comp_ () const { return comp; } bool push_back (int); bool push_front (int); void expand (); void expand (int); void clear (); const string at (int); int operator[ ](int); Vector& operator+ (Vector&); Vector& operator- (const Vector&); bool operator== (const Vector&); bool operator!= (const Vector&); ~Vector() { delete [] data; } }; ostream& operator<< (ostream&, const Vector&); #endif Vector.cpp #include <iostream> #include <string> #include "Vector.h" using namespace std; const string Vector::at(int i) { this[i]; } void Vector::expand() { expand(size); } void Vector::expand(int n ) { int * newdata = new int [comp * 2]; if (*data != NULL) { for (int i = 0; i <= (comp); i++) { newdata[i] = data[i]; } newdata -= comp; comp += n; delete [] data; *data = *newdata; } else if ( *data == NULL || comp == 0) { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; size = 0; } } bool Vector::push_back(int n) { if (comp = 0) { expand(); } for (int k = 0; k != 2; k++) { if ( size != comp ){ data[size] = n; size++; return true; } else { expand(); } } return false; } void Vector::clear() { delete [] data; comp = 0; size = 0; } int Vector::operator[] (int place) { return (data[place]); } Vector& Vector::operator+ (Vector& n) { int temp_int = 0; if (size > n.size_() || size == n.size_()) { temp_int = size; } else if (size < n.size_()) { temp_int = n.size_(); } Vector newone(temp_int); int temp_2_int = 0; for ( int j = 0; j <= temp_int && j <= n.size_() && j <= size; j++) { temp_2_int = n[j] + data[j]; newone[j] = temp_2_int; } //////////////////////////////////////////////////////////// return newone; //////////////////////////////////////////////////////////// } ostream& operator<< (ostream& out, const Vector& n) { for (int i = 0; i <= n.size_(); i++) { //////////////////////////////////////////////////////////// out << n[i] << " "; //////////////////////////////////////////////////////////// } return out; } Errors: out << n[i] << " "; error C2678: binary '[' : no operator found which takes a left-hand operand of type 'const Vector' (or there is no acceptable conversion) return newone; error C2106: '=' : left operand must be l-value As stated above, I am a student going into Computer Science as my selected major I would appreciate tips, pointers, and better ways to do stuff :D

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  • using compareTo in Binary Search Tree program

    - by Scott Rogener
    I've been working on this program for a few days now and I've implemented a few of the primary methods in my BinarySearchTree class such as insert and delete. Insert seemed to be working fine, but once I try to delete I kept getting errors. So after playing around with the code I wanted to test my compareTo methods. I created two new nodes and tried to compare them and I get this error: Exception in thread "main" java.lang.ClassCastException: TreeNode cannot be cast to java.lang.Integer at java.lang.Integer.compareTo(Unknown Source) at TreeNode.compareTo(TreeNode.java:16) at BinarySearchTree.myComparision(BinarySearchTree.java:177) at main.main(main.java:14) Here is my class for creating the nodes: public class TreeNode<T> implements Comparable { protected TreeNode<T> left, right; protected Object element; public TreeNode(Object obj) { element=obj; left=null; right=null; } public int compareTo(Object node) { return ((Comparable) this.element).compareTo(node); } } Am I doing the compareTo method all wrong? I would like to create trees that can handle integers and strings (seperatly of course)

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  • Using recursion to to trim a binary tree based on a given min and max value

    - by Justin
    As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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