Search Results

Search found 19480 results on 780 pages for 'do your own homework'.

Page 37/780 | < Previous Page | 33 34 35 36 37 38 39 40 41 42 43 44  | Next Page >

  • Problem passing strings with PHP post

    - by andy
    Hi Guys, Basically I'm developing a (very) simple search engine. You can enter a search term and you are taken to the results page - which works fine. However on the results page, I have a button that will take you to the next 10 results: where $term is the $_POST['term'] value. echo "<input type='hidden' name='term' value='" . $term . "'>"; This causes the following problem with the term is for example "aidan's". When the next 10 button is clicked, the term becomes aidan\ and no further results are found. I am not doing anything to $term. I usually use Java, but am required to use PHP for this uni assignment! Any help would be greatly appreciated.

    Read the article

  • Virtual Memory and Paging

    - by Kenshin
    Hello, I am doing some exercices to understand how the virtual memory and paging works, my question is as follows : Suppose we use a paged memory with pages of 1024 bytes, the virtual address space is of 8 pages but the physical memory can only contain 4 frames of pages. Replacement policy is LRU. What is the physical address in main memory that corresponds to virtual address 4096? and how do you get to that result? Same thing as question 1 but with virtual address 1024 A page fault occurs when accessing a word in page 0, which page frame will be used to receive the virtual page 0? Page Image

    Read the article

  • How do you display a binary search tree?

    - by fakeit
    I'm being asked to display a binary search tree in sorted order. The nodes of the tree contain strings. I'm not exactly sure what the best way is to attack this problem. Should I be traversing the tree and displaying as I go? Should I flatten the tree into an array and then use a sorting algorithm before I display? I'm not looking for the actual code, just a guide where to go next.

    Read the article

  • Theory of computation - Using the pumping lemma for context free languages

    - 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?

    Read the article

  • help me to my project [closed]

    - by latha
    hi. plz tel me to how to collect data for my project. as my projct relate to "sustainable urban infrastructure" so help me that what what data should i put to my project.i dont had any idea that how to do vit data analysis .so please direct me .

    Read the article

  • How do I create a list or set object in a class in Python?

    - by Az
    For my project, the role of the Lecturer (defined as a class) is to offer projects to students. Project itself is also a class. I have some global dictionaries, keyed by the unique numeric id's for lecturers and projects that map to objects. Thus for the "lecturers" dictionary (currently): lecturer[id] = Lecturer(lec_name, lec_id, max_students) I'm currently reading in a white-space delimited text file that has been generated from a database. I have no direct access to the database so I haven't much say on how the file is formatted. Here's a fictionalised snippet that shows how the text file is structured. Please pardon the cheesiness. 0001 001 "Miyamoto, S." "Even Newer Super Mario Bros" 0002 001 "Miyamoto, S." "Legend of Zelda: Skies of Hyrule" 0003 002 "Molyneux, P." "Project Milo" 0004 002 "Molyneux, P." "Fable III" 0005 003 "Blow, J." "Ponytail" The structure of each line is basically proj_id, lec_id, lec_name, proj_name. Now, I'm currently reading the relevant data into the relevant objects. Thus, proj_id is stored in class Project whereas lec_name is a class Lecturer object, et al. The Lecturer and Project classes are not currently related. However, as I read in each line from the text file, for that line, I wish to read in the project offered by the lecturer into the Lecturer class; I'm already reading the proj_id into the Project class. I'd like to create an object in Lecturer called offered_proj which should be a set or list of the projects offered by that lecturer. Thus whenever, for a line, I read in a new project under the same lec_id, offered_proj will be updated with that project. If I wanted to get display a list of projects offered by a lecturer I'd ideally just want to use print lecturers[lec_id].offered_proj. My Python isn't great and I'd appreciate it if someone could show me a way to do that. I'm not sure if it's better as a set or a list, as well. Update After the advice from Alex Martelli and Oddthinking I went back and made some changes and tried to print the results. Here's the code snippet: for line in csv_file: proj_id = int(line[0]) lec_id = int(line[1]) lec_name = line[2] proj_name = line[3] projects[proj_id] = Project(proj_id, proj_name) lecturers[lec_id] = Lecturer(lec_id, lec_name) if lec_id in lecturers.keys(): lecturers[lec_id].offered_proj.add(proj_id) print lec_id, lecturers[lec_id].offered_proj The print lecturers[lec_id].offered_proj line prints the following output: 001 set([0001]) 001 set([0002]) 002 set([0003]) 002 set([0004]) 003 set([0005]) It basically feels like the set is being over-written or somesuch. So if I try to print for a specific lecturer print lec_id, lecturers[001].offered_proj all I get is the last the proj_id that has been read in.

    Read the article

  • Intel Assembly Programming

    - by Kay
    class MyString{ char buf[100]; int len; boolean append(MyString str){ int k; if(this.len + str.len>100){ for(k=0; k<str.len; k++){ this.buf[this.len] = str.buf[k]; this.len ++; } return false; } return true; } } Does the above translate to: start: push ebp ; save calling ebp mov ebp, esp ; setup new ebp push esi ; push ebx ; mov esi, [ebp + 8] ; esi = 'this' mov ebx, [ebp + 14] ; ebx = str mov ecx, 0 ; k=0 mov edx, [esi + 200] ; edx = this.len append: cmp edx + [ebx + 200], 100 jle ret_true ; if (this.len + str.len)<= 100 then ret_true cmp ecx, edx jge ret_false ; if k >= str.len then ret_false mov [esi + edx], [ebx + 2*ecx] ; this.buf[this.len] = str.buf[k] inc edx ; this.len++ aux: inc ecx ; k++ jmp append ret_true: pop ebx ; restore ebx pop esi ; restore esi pop ebp ; restore ebp ret true ret_false: pop ebx ; restore ebx pop esi ; restore esi pop ebp ; restore ebp ret false My greatest difficulty here is figuring out what to push onto the stack and the math for pointers. NOTE: I'm not allowed to use global variables and i must assume 32-bit ints, 16-bit chars and 8-bit booleans.

    Read the article

  • 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.

    Read the article

  • Limit Connections with semaphores

    - by Robert
    I'm trying to limit the number of connections my server will accept using semaphores, but when running, my code doesn't seem to make this restriction - am I using the semaphore correctly? eg. I have hardcoded the number of permit as 2, but I can connect an unlimited number of clients... public class EServer implements Runnable { private ServerSocket serverSocket; private int numberofConnections = 0; private Semaphore sem = new Semaphore(2); private volatile boolean keepProcessing = true; public EServer(int port) throws IOException { serverSocket = new ServerSocket(port); } @Override public void run() { while (keepProcessing) { try { sem.acquire(); Socket socket = serverSocket.accept(); process(socket, getNextConnectionNumber()); } catch (Exception e) { } finally { sem.release(); } } closeIgnoringException(serverSocket); } private synchronized int getNextConnectionNumber() { return ++numberofConnections; } // processing related methods }

    Read the article

  • 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!

    Read the article

  • Spatial domain to frequency domain

    - by John Elway
    I know about Fourier Transforms, but I don't know how to apply it here, and I think that is over the top. I gave my ideas of the responses, but I really don't know what I'm looking for... Supposed that you form a low-pass spatial filter h(x,y) that averages all the eight immediate neighbors of a pixel (x,y) but excludes itself. a. Find the equivalent frequency domain filter H(u,v): My answer is to (a): 1/8*H(u-1, v-1) + 1/8*H(u-1, v) + 1/8*H(u-1, v+1) + 1/8*H(u, v-1) + 0 + 1/8*H(u, v+1) + 1/8*H(u+1, v-1) + 1/8*H(u+1, v) + 1/8*H(u-1, v-1) is this the frequency domain? b. Show that your result is again a low-pass filter. does this have to do with the coefficients being positive?

    Read the article

  • assembly language programming (prime number)

    - by chris
    Prompt the user for a positive three digit number, then read it. Let's call it N. Divide into N all integer values from 2 to (N/2)+1 and test to see if the division was even, in which case N is instantly shown to be non-prime. Output a message printing N and saying that it is not prime. If none of those integer values divide evenly (remainder never is zero), then N is shown to be prime. Output a message printing N and saying that it is prime. Ask the user if he or she wants to test another number; if the user types "n" or "N", quit. If "y" or "Y", jump back and repeat. Comments in your code are essential. Hi. I am kinda in rush to do this.. please help me doing it. I'll be much appreciated. thank you

    Read the article

  • Returning new object, overwrite the existing one in Java

    - by lupin
    Note: This is an assignment. Hi, Ok I have this method that will create a supposedly union of 2 sets. i mport java.io.*; class Set { public int numberOfElements; public String[] setElements; public int maxNumberOfElements; // constructor for our Set class public Set(int numberOfE, int setE, int maxNumberOfE) { this.numberOfElements = numberOfE; this.setElements = new String[setE]; this.maxNumberOfElements = maxNumberOfE; } // Helper method to shorten/remove element of array since we're using basic array instead of ArrayList or HashSet from collection interface :( static String[] removeAt(int k, String[] arr) { final int L = arr.length; String[] ret = new String[L - 1]; System.arraycopy(arr, 0, ret, 0, k); System.arraycopy(arr, k + 1, ret, k, L - k - 1); return ret; } int findElement(String element) { int retval = 0; for ( int i = 0; i < setElements.length; i++) { if ( setElements[i] != null && setElements[i].equals(element) ) { return retval = i; } retval = -1; } return retval; } void add(String newValue) { int elem = findElement(newValue); if( numberOfElements < maxNumberOfElements && elem == -1 ) { setElements[numberOfElements] = newValue; numberOfElements++; } } int getLength() { if ( setElements != null ) { return setElements.length; } else { return 0; } } String[] emptySet() { setElements = new String[0]; return setElements; } Boolean isFull() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == maxNumberOfElements ){ return True; } else { return False; } } Boolean isEmpty() { Boolean True = new Boolean(true); Boolean False = new Boolean(false); if ( setElements.length == 0 ) { return True; } else { return False; } } void remove(String newValue) { for ( int i = 0; i < setElements.length; i++) { if ( setElements[i] != null && setElements[i].equals(newValue) ) { setElements = removeAt(i,setElements); } } } int isAMember(String element) { int retval = -1; for ( int i = 0; i < setElements.length; i++ ) { if (setElements[i] != null && setElements[i].equals(element)) { return retval = i; } } return retval; } void printSet() { for ( int i = 0; i < setElements.length; i++) { if (setElements[i] != null) { System.out.println("Member elements on index: "+ i +" " + setElements[i]); } } } String[] getMember() { String[] tempArray = new String[setElements.length]; for ( int i = 0; i < setElements.length; i++) { if(setElements[i] != null) { tempArray[i] = setElements[i]; } } return tempArray; } Set union(Set x, Set y) { String[] newXtemparray = new String[x.getLength()]; String[] newYtemparray = new String[y.getLength()]; int len = newYtemparray.length + newXtemparray.length; Set temp = new Set(0,len,len); newXtemparray = x.getMember(); newYtemparray = x.getMember(); for(int i = 0; i < newYtemparray.length; i++) { temp.add(newYtemparray[i]); } for(int j = 0; j < newXtemparray.length; j++) { temp.add(newXtemparray[j]); } return temp; } Set difference(Set x, Set y) { String[] newXtemparray = new String[x.getLength()]; String[] newYtemparray = new String[y.getLength()]; int len = newYtemparray.length + newXtemparray.length; Set temp = new Set(0,len,len); newXtemparray = x.getMember(); newYtemparray = x.getMember(); for(int i = 0; i < newXtemparray.length; i++) { temp.add(newYtemparray[i]); } for(int j = 0; j < newYtemparray.length; j++) { int retval = temp.findElement(newYtemparray[j]); if( retval != -1 ) { temp.remove(newYtemparray[j]); } } return temp; } } // This is the SetDemo class that will make use of our Set class class SetDemo { public static void main(String[] args) { //get input from keyboard BufferedReader keyboard; InputStreamReader reader; String temp = ""; reader = new InputStreamReader(System.in); keyboard = new BufferedReader(reader); try { System.out.println("Enter string element to be added" ); temp = keyboard.readLine( ); System.out.println("You entered " + temp ); } catch (IOException IOerr) { System.out.println("There was an error during input"); } /* ************************************************************************** * Test cases for our new created Set class. * ************************************************************************** */ Set setA = new Set(0,10,10); setA.add(temp); setA.add("b"); setA.add("b"); setA.add("hello"); setA.add("world"); setA.add("six"); setA.add("seven"); setA.add("b"); int size = setA.getLength(); System.out.println("Set size is: " + size ); Boolean isempty = setA.isEmpty(); System.out.println("Set is empty? " + isempty ); int ismember = setA.isAMember("sixb"); System.out.println("Element sixb is member of setA? " + ismember ); Boolean output = setA.isFull(); System.out.println("Set is full? " + output ); //setA.printSet(); int index = setA.findElement("world"); System.out.println("Element b located on index: " + index ); setA.remove("b"); //setA.emptySet(); int resize = setA.getLength(); System.out.println("Set size is: " + resize ); //setA.printSet(); Set setB = new Set(0,10,10); setB.add("b"); setB.add("z"); setB.add("x"); setB.add("y"); Set setC = setA.union(setB,setA); System.out.println("Elements of setA"); setA.printSet(); System.out.println("Union of setA and setB"); setC.printSet(); } } The union method works a sense that somehow I can call another method on it but it doesn't do the job, i supposedly would create and union of all elements of setA and setB but it only return element of setB. Sample output follows: java SetDemo Enter string element to be added hello You entered hello Set size is: 10 Set is empty? false Element sixb is member of setA? -1 Set is full? true Element b located on index: 2 Set size is: 9 Elements of setA Member elements on index: 0 hello Member elements on index: 1 world Member elements on index: 2 six Member elements on index: 3 seven Union of setA and setB Member elements on index: 0 b Member elements on index: 1 z Member elements on index: 2 x Member elements on index: 3 y thanks, lupin

    Read the article

  • Reliable UDP

    - by suresh
    How can I develop a Linux kernel module in order to make UDP reliable? This is my college assignment and I don't how to proceed. how to do change the default UDP behaviour in linux kernel by loading a new kernel module? and how to program such kernel module?

    Read the article

  • Checking a set of listbox items against a text box vb.net

    - by Shane Fagan
    Hi all I have this code to check if an item from a textbox is in a listbox and its giving me the error at the bottom. Any ideas what im doing wrong? I copied it from another part of my project and it was working for that part so I cant see whats wrong. If LocationsSearchTextBox.Text <> "" And LocationListBox.Items.Count > 0 Then tempInt = 0 While (tempInt < ClientListBox.Items.Count) If LocationListBox.Items(tempInt).ToString.Contains(LocationsSearchTextBox.Text) = False Then LocationListBox.Items.RemoveAt(tempInt) End If tempInt += 1 End While End If System.ArgumentOutOfRangeException was unhandled Message="InvalidArgument=Value of '2' is not valid for 'index'. Parameter name: index" ParamName="index" Source="System.Windows.Forms" StackTrace: at System.Windows.Forms.ListBox.ObjectCollection.get_Item(Int32 index) at AuctioneerProject.Viewing.LocationsSearchTextBox_KeyPress(Object sender, KeyPressEventArgs e) in C:\Users\admin\Desktop\Auctioneers\AuctioneerProject\AuctioneerProject\Viewing.vb:line 301 at System.Windows.Forms.Control.OnKeyPress(KeyPressEventArgs e) at System.Windows.Forms.Control.ProcessKeyEventArgs(Message& m) at System.Windows.Forms.Control.ProcessKeyMessage(Message& m) at System.Windows.Forms.Control.WmKeyChar(Message& m) at System.Windows.Forms.Control.WndProc(Message& m) at System.Windows.Forms.TextBoxBase.WndProc(Message& m) at System.Windows.Forms.TextBox.WndProc(Message& m) at System.Windows.Forms.Control.ControlNativeWindow.OnMessage(Message& m) at System.Windows.Forms.Control.ControlNativeWindow.WndProc(Message& m) at System.Windows.Forms.NativeWindow.DebuggableCallback(IntPtr hWnd, Int32 msg, IntPtr wparam, IntPtr lparam) at System.Windows.Forms.UnsafeNativeMethods.DispatchMessageW(MSG& msg) at System.Windows.Forms.Application.ComponentManager.System.Windows.Forms.UnsafeNativeMethods.IMsoComponentManager.FPushMessageLoop(Int32 dwComponentID, Int32 reason, Int32 pvLoopData) at System.Windows.Forms.Application.ThreadContext.RunMessageLoopInner(Int32 reason, ApplicationContext context) at System.Windows.Forms.Application.ThreadContext.RunMessageLoop(Int32 reason, ApplicationContext context) at System.Windows.Forms.Application.Run(ApplicationContext context) at Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase.OnRun() at Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase.DoApplicationModel() at Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase.Run(String[] commandLine) at AuctioneerProject.My.MyApplication.Main(String[] Args) in 17d14f5c-a337-4978-8281-53493378c1071.vb:line 81 at System.AppDomain._nExecuteAssembly(Assembly assembly, String[] args) at System.AppDomain.ExecuteAssembly(String assemblyFile, Evidence assemblySecurity, String[] args) at Microsoft.VisualStudio.HostingProcess.HostProc.RunUsersAssembly() at System.Threading.ThreadHelper.ThreadStart_Context(Object state) at System.Threading.ExecutionContext.Run(ExecutionContext executionContext, ContextCallback callback, Object state) at System.Threading.ThreadHelper.ThreadStart() InnerException:

    Read the article

  • 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???

    Read the article

< Previous Page | 33 34 35 36 37 38 39 40 41 42 43 44  | Next Page >