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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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  • 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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  • 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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  • 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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  • 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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  • Simplest way to automatically alter "const" value in Java during compile time

    - by Michael Mao
    Hi all: This is a question corresponds to my uni assignment so I am very sorry to say I cannot adopt any one of the following best practices in a short time -- you know -- assignment is due tomorrow :( link to Best way to alter const in Java on StackOverflow Basically the only task (I hope so) left for me is the performance tuning. I've got a bunch of predefined "const" values in my single-class agent source code like this: //static final values private static final long FT_THRESHOLD = 400; private static final long FT_THRESHOLD_MARGIN = 50; private static final long FT_SMOOTH_PRICE_IDICATOR = 20; private static final long FT_BUY_PRICE_MODIFIER = 0; private static final long FT_LAST_ROUNDS_STARTTIME = 90; private static final long FT_AMOUNT_ADJUST_MODIFIER = 5; private static final long FT_HISTORY_PIRCES_LENGTH = 10; private static final long FT_TRACK_DURATION = 5; private static final int MAX_BED_BID_NUM_PER_AUC = 12; I can definitely alter the values manually and then compile the code to give it another go around. But the execution time for a thorough "statistic analysis" usually requires over 2000 times of execution, which will typically lasts more than half an hour on my own laptop... So I hope there is a way to alter values using other ways than dig into the source code to change the "const" values there, so I can automatically distributed compiled code to other people's PC and let them run the statistic analysis instead. One other reason for a automatically value adjustment is that I can try using my own agent to defeat itself by choosing different "const" values. Although my values are derived from previous history and statistical results, they are far from the most optimized values. I hope there is a easy way so I can quickly adopt that so to have a good sleep tonight while the computer does everything for me... :) Any hints on this sort of stuff? Any suggestion is welcomed and much appreciated.

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  • Regular Expression question

    - by Mohammad Kotb
    Hi, In my academic assignment, I want make a regular expression to match a word with the following specifications: word length greater than or equal 1 and less than or equal 8 contains letters, digits, and underscore first digit is a letter only word is not A,X,S,T or PC,SW I tried for this regex but can't continue (My big problem is to make the word not equal to PC and SW) ([a-zA-Z&&[^AXST]])|([a-zA-Z][\w]{0,7}) But in the previous regex I didn't handle the that it is not PC and SW Thanks,

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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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  • Theory of computation - Using the pumping lemma for CFLs

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

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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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  • Log 2 N generic comparison tree

    - by Morano88
    Hey! I'm working on an algorithm for Redundant Binary Representation (RBR) where every two bits represent a digit. I designed the comparator that takes 4 bits and gives out 2 bits. I want to make the comparison in log 2 n so If I have X and Y .. I compare every 2 bits of X with every 2 bits of Y. This is smooth if the number of bits of X or Y equals n where (n = 2^X) i.e n = 2,4,8,16,32,... etc. Like this : However, If my input let us say is 6 or 10 .. then it becomes not smooth and I have to take into account some odd situations like this : I have a shallow experience in algorithms .. is there a generic way to do it .. so at the end I get only 2 bits no matter what the input is ? I just need hints or pseudo-code. If my question is not appropriate here .. so feel free to flag it or tell me to remove it. I'm using VHDL by the way!

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  • Write a program that allows the user to enter a string and then prints the letters of the String sep

    - by WM
    The output is always a String, for example H,E,L,L,O,. How could I limit the commas? I want the commas only between letters, for example H,E,L,L,O. import java.util.Scanner; import java.lang.String; public class forLoop { public static void main(String[] args) { Scanner Scan = new Scanner(System.in); System.out.print("Enter a string: "); String Str1 = Scan.next(); String newString=""; String Str2 =""; for (int i=0; i < Str1.length(); i++) { newString = Str1.charAt(i) + ","; Str2 = Str2 + newString; } System.out.print(Str2); } }

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  • Dynamic programming: Largest diamond(rhombus) block

    - by Darksody
    I have a small program to do in Java. I have a 2D array filled with 0 and 1, and I must find the largest rhombus (as in square rotated by 90 degrees) and their numbers. Example: 0 1 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Result: 1 1 1 1 1 1 1 1 1 1 1 1 1 The problem is similar to this SO question. If you have any idea, post it here.

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  • Unknown error in Producer/Consumer program, believe it to be an infinite loop.

    - by ray2k
    Hello, I am writing a program that is solving the producer/consumer problem, specifically the bounded-buffer version(i believe they mean the same thing). The producer will be generating x number of random numbers, where x is a command line parameter to my program. At the current moment, I believe my program is entering an infinite loop, but I'm not sure why it is occurring. I believe I am executing the semaphores correctly. You compile it like this: gcc -o prodcon prodcon.cpp -lpthread -lrt Then to run, ./prodcon 100(the number of randum nums to produce) This is my code. typedef int buffer_item; #include <stdlib.h> #include <stdio.h> #include <pthread.h> #include <semaphore.h> #include <unistd.h> #define BUFF_SIZE 10 #define RAND_DIVISOR 100000000 #define TRUE 1 //two threads void *Producer(void *param); void *Consumer(void *param); int insert_item(buffer_item item); int remove_item(buffer_item *item); int returnRandom(); //the global semaphores sem_t empty, full, mutex; //the buffer buffer_item buf[BUFF_SIZE]; //buffer counter int counter; //number of random numbers to produce int numRand; int main(int argc, char** argv) { /* thread ids and attributes */ pthread_t pid, cid; pthread_attr_t attr; pthread_attr_init(&attr); pthread_attr_setscope(&attr, PTHREAD_SCOPE_SYSTEM); numRand = atoi(argv[1]); sem_init(&empty,0,BUFF_SIZE); sem_init(&full,0,0); sem_init(&mutex,0,0); printf("main started\n"); pthread_create(&pid, &attr, Producer, NULL); pthread_create(&cid, &attr, Consumer, NULL); printf("main gets here"); pthread_join(pid, NULL); pthread_join(cid, NULL); printf("main done\n"); return 0; } //generates a randum number between 1 and 100 int returnRandom() { int num; srand(time(NULL)); num = rand() % 100 + 1; return num; } //begin producing items void *Producer(void *param) { buffer_item item; int i; for(i = 0; i < numRand; i++) { //sleep for a random period of time int rNum = rand() / RAND_DIVISOR; sleep(rNum); //generate a random number item = returnRandom(); //acquire the empty lock sem_wait(&empty); //acquire the mutex lock sem_wait(&mutex); if(insert_item(item)) { fprintf(stderr, " Producer report error condition\n"); } else { printf("producer produced %d\n", item); } /* release the mutex lock */ sem_post(&mutex); /* signal full */ sem_post(&full); } return NULL; } /* Consumer Thread */ void *Consumer(void *param) { buffer_item item; int i; for(i = 0; i < numRand; i++) { /* sleep for a random period of time */ int rNum = rand() / RAND_DIVISOR; sleep(rNum); /* aquire the full lock */ sem_wait(&full); /* aquire the mutex lock */ sem_wait(&mutex); if(remove_item(&item)) { fprintf(stderr, "Consumer report error condition\n"); } else { printf("consumer consumed %d\n", item); } /* release the mutex lock */ sem_post(&mutex); /* signal empty */ sem_post(&empty); } return NULL; } /* Add an item to the buffer */ int insert_item(buffer_item item) { /* When the buffer is not full add the item and increment the counter*/ if(counter < BUFF_SIZE) { buf[counter] = item; counter++; return 0; } else { /* Error the buffer is full */ return -1; } } /* Remove an item from the buffer */ int remove_item(buffer_item *item) { /* When the buffer is not empty remove the item and decrement the counter */ if(counter > 0) { *item = buf[(counter-1)]; counter--; return 0; } else { /* Error buffer empty */ return -1; } }

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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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  • How do I get the next token in a Cstring if I want to use it as an int? (c++)

    - by Van
    My objective is to take directions from a user and eventually a text file to move a robot. The catch is that I must use Cstrings(such as char word[];) rather than the std::string and tokenize them for use. the code looks like this: void Navigator::manualDrive() { char uinput[1]; char delim[] = " "; char *token; cout << "Enter your directions below: \n"; cin.ignore(); cin.getline (uinput, 256); token=strtok(uinput, delim); if(token == "forward") { int inches; inches=token+1; travel(inches); } } I've never used Cstrings I've never tokenized anything before, and I don't know how to write this. Our T.A.'s expect us to google and find all the answers because they are aware we've never been taught these methods. Everyone in my lab is having much more trouble than usual. I don't know the code to write but I know what I want my program to do. I want it to execute like this: 1) Ask for directions. 2) cin.getline the users input 3) tokenize the inputed string 4) if the first word token == "forward" move to the next token and find out how many inches to move forward then move forward 5) else if the first token == "turn" move to the next token. if the next token == "left" move to the next token and find out how many degrees to turn left I will have to do this for forward x, backward x, turn left x, turn right x, and stop(where x is in inches or degrees). I already wrote functions that tell the robot how to move forward an inch and turn in degrees. I just need to know how to convert the inputted strings to all lowercase letters and move from token to token and convert or extract the numbers from the string to use them as integers. If all is not clear you can read my lab write up at this link: http://www.cs.utk.edu/~cs102/robot_labs/Lab9.html If anything is unclear please let me know, and I will clarify as best I can.

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  • A specific string format with a number and character together represeting a certain item

    - by sil3nt
    Hello there, I have a string which looks like this "a 3e,6s,1d,3g,22r,7c 3g,5r,9c 19.3", how do I go through it and extract the integers and assign them to its corresponding letter variable?. (i have integer variables d,r,e,g,s and c). The first letter in the string represents a function, "3e,6s,1d,3g,22r,7c" and "3g,5r,9c" are two separate containers . And the last decimal value represents a number which needs to be broken down into those variable numbers. my problem is extracting those integers with the letters after it and assigning them into there corresponding letter. and any number with a negative sign or a space in between the number and the letter is invalid. How on earth do i do this?

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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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  • How to draw the "trail" in a maze solving application

    - by snow-spur
    Hello i have designed a maze and i want to draw a path between the cells as the 'person' moves from one cell to the next. So each time i move the cell a line is drawn Also i am using the graphics module The graphics module is an object oriented library Im importing from graphics import* from maze import* my circle which is my cell center = Point(15, 15) c = Circle(center, 12) c.setFill('blue') c.setOutline('yellow') c.draw(win) p1 = Point(c.getCenter().getX(), c.getCenter().getY()) this is my loop if mazez.blockedCount(cloc)> 2: mazez.addDecoration(cloc, "grey") mazez[cloc].deadend = True c.move(-25, 0) p2 = Point(getX(), getY()) line = graphics.Line(p1, p2) cloc.col = cloc.col - 1 Now it says getX not defined every time i press a key is this because of p2???

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