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  • C++ design question on traversing binary trees

    - by user231536
    I have a binary tree T which I would like to copy to another tree. Suppose I have a visit method that gets evaluated at every node: struct visit { virtual void operator() (node* n)=0; }; and I have a visitor algorithm void visitor(node* t, visit& v) { //do a preorder traversal using stack or recursion if (!t) return; v(t); visitor(t->left, v); visitor(t->right, v); } I have 2 questions: I settled on using the functor based approach because I see that boost graph does this (vertex visitors). Also I tend to repeat the same code to traverse the tree and do different things at each node. Is this a good design to get rid of duplicated code? What other alternative designs are there? How do I use this to create a new binary tree from an existing one? I can keep a stack on the visit functor if I want, but it gets tied to the algorithm in visitor. How would I incorporate postorder traversals here ? Another functor class?

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  • check if a tree is a binary search tree

    - by TimeToCodeTheRoad
    I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Pair p{ boolean isTrue; int min; int max; } public boo lean isBst(BNode v){ return isBST1(v).isTrue; } public Pair isBST1(BNode v){ if(v==null) return new Pair(true, INTEGER.MIN,INTEGER.MAX); if(v.left==null && v.right==null) return new Pair(true, v.data, v.data); Pair pLeft=isBST1(v.left); Pair pRight=isBST1(v.right); boolean check=pLeft.max<v.data && v.data<= pRight.min; Pair p=new Pair(); p.isTrue=check&&pLeft.isTrue&&pRight.isTrue; p.min=pLeft.min; p.max=pRight.max; return p; } Note: This function checks if a tree is a binary search tree

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  • Getting zeros between data while reading a binary file in C

    - by indiajoe
    I have a binary data which I am reading into an array of long integers using a C programme. hexdump of the binary data shows, that after first few data points , it starts again at a location 20000 hexa adresses away. hexdump output is as shown below. 0000000 0000 0000 0000 0000 0000 0000 0000 0000 * 0020000 0000 0000 0053 0000 0064 0000 006b 0000 0020010 0066 0000 0068 0000 0066 0000 005d 0000 0020020 0087 0000 0059 0000 0062 0000 0066 0000 ........ and so on... But when I read it into an array 'data' of long integers. by the typical fread command fread(data,sizeof(*data),filelength/sizeof(*data),fd); It is filling up with all zeros in my data array till it reaches the 20000 location. After that it reads in data correctly. Why is it reading regions where my file is not there? Or how will I make it read only my file, not anything inbetween which are not in file? I know it looks like a trivial problem, but I cannot figure it out even after googling one night.. Can anyone suggest me where I am doing it wrong? Other Info : I am working on a gnu/linux machine. (slax-atma distro to be specific) My C compiler is gcc.

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  • How to serve static files for multiple Django projects via nginx to same domain

    - by thanley
    I am trying to setup my nginx conf so that I can serve the relevant files for my multiple Django projects. Ultimately I want each app to be available at www.example.com/app1, www.example.com/app2 etc. They all serve static files from a 'static-files' directory located in their respective project root. The project structure: Home Ubuntu Web www.example.com ref logs app app1 app1 static bower_components templatetags app1_project templates static-files app2 app2 static templates templatetags app2_project static-files app3 tests templates static-files static app3_project app3 venv When I use the conf below, there are no problems for serving the static-files for the app that I designate in the /static/ location. I can also access the different apps found at their locations. However, I cannot figure out how to serve all of the static files for all the apps at the same time. I have looked into using the 'try_files' command for the static location, but cannot figure out how to see if it is working or not. Nginx Conf - Only serving static files for one app: server { listen 80; server_name example.com; server_name www.example.com; access_log /home/ubuntu/web/www.example.com/logs/access.log; error_log /home/ubuntu/web/www.example.com/logs/error.log; root /home/ubuntu/web/www.example.com/; location /static/ { alias /home/ubuntu/web/www.example.com/app/app1/static-files/; } location /media/ { alias /home/ubuntu/web/www.example.com/media/; } location /app1/ { include uwsgi_params; uwsgi_param SCRIPT_NAME /app1; uwsgi_modifier1 30; uwsgi_pass unix:///home/ubuntu/web/www.example.com/app1.sock; } location /app2/ { include uwsgi_params; uwsgi_param SCRIPT_NAME /app2; uwsgi_modifier1 30; uwsgi_pass unix:///home/ubuntu/web/www.example.com/app2.sock; } location /app3/ { include uwsgi_params; uwsgi_param SCRIPT_NAME /app3; uwsgi_modifier1 30; uwsgi_pass unix:///home/ubuntu/web/www.example.com/app3.sock; } # what to serve if upstream is not available or crashes error_page 400 /static/400.html; error_page 403 /static/403.html; error_page 404 /static/404.html; error_page 500 502 503 504 /static/500.html; # Compression gzip on; gzip_http_version 1.0; gzip_comp_level 5; gzip_proxied any; gzip_min_length 1100; gzip_buffers 16 8k; gzip_types text/plain text/css application/x-javascript text/xml application/xml application/xml+rss text/javascript; # Some version of IE 6 don't handle compression well on some mime-types, # so just disable for them gzip_disable "MSIE [1-6].(?!.*SV1)"; # Set a vary header so downstream proxies don't send cached gzipped # content to IE6 gzip_vary on; } Essentially I want to have something like (I know this won't work) location /static/ { alias /home/ubuntu/web/www.example.com/app/app1/static-files/; alias /home/ubuntu/web/www.example.com/app/app2/static-files/; alias /home/ubuntu/web/www.example.com/app/app3/static-files/; } or (where it can serve the static files based on the uri) location /static/ { try_files $uri $uri/ =404; } So basically, if I use try_files like above, is the problem in my project directory structure? Or am I totally off base on this and I need to put each app in a subdomain instead of going this route? Thanks for any suggestions TLDR: I want to go to: www.example.com/APP_NAME_HERE And have nginx serve the static location: /home/ubuntu/web/www.example.com/app/APP_NAME_HERE/static-files/;

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  • C++ exceptions binary compatibility

    - by aaa
    hi. my project uses 2 different C++ compilers, g++ and nvcc (cuda compiler). I have noticed exception thrown from nvcc object files are not caught in g++ object files. are C++ exceptions supposed to be binary compatible in the same machine? what can cause such behavior?

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  • Binary Search Tree Implementation

    - by Gabe
    I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

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  • Parsing a given binary tree using python?

    - by kaushik
    Parse a binary tree,referring to given set of features,answering decision tree question at each node to decide left child or right child and find the path to leaf node according to answer given to the decision tree.. input wil be a set of feature which wil help in answering the question at each level to choose the left or right half and the output will be the leaf node.. i need help in implementing this can anyone suggest methods?? Please answer... thanks in advance..

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  • C# TCP socket and binary data

    - by MD
    Hi @All How to send binary data (01110110 for exemple) with C# throught a TCP (using SSL) socket ? I'm using : SslStream.Write() and h[0] = (byte)Convert.ToByte("01110110"); isn't working Thanks.

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  • C syntax or binary optimized syntax?

    - by Dpp
    Let's take an simple example of two lines supposedly doing the same thing: if (value = 96 || value < 0) ... or if (value & ~ 95) ... Say 'If's are costly in a loop of thousands of iterations, is it better to keep with the traditional C syntax or better to find a binary optimized one if possible?

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  • convert a binary file in a list (python)

    - by beratch
    Hi all, I'd like to be able to open a binary file, and make a list (kind of array) with all the chars in, like : "\x21\x23\x22\x21\x22\x31" to ["\x21","\x23","\x22","\x21","\x22","\x31"] What would be the best solution to convert it ? Thanks !

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  • Write binary stream to browser using PHP

    - by Dave Jarvis
    Background Trying to stream a PDF report written using iReport through PHP to the browser. The general problem is: how do you write binary data to the browser using PHP? Working Code The following code does the job, but (for many reasons) it is not as efficient as it should be (the code writes a file then sends the file contents the browser). // Load the MySQL database driver. // java( 'java.lang.Class' )->forName( 'com.mysql.jdbc.Driver' ); // Attempt a database connection. // $conn = java( 'java.sql.DriverManager' )->getConnection( "jdbc:mysql://localhost:3306/climate?user=$user&password=$password" ); // Extract parameters. // $params = new java('java.util.HashMap'); $params->put('DistrictCode', '101'); $params->put('StationCode', '0066'); $params->put('CategoryCode', '010'); // Use the fill manager to produce the report. // $fm = java('net.sf.jasperreports.engine.JasperFillManager'); $pm = $fm->fillReport($report, $params, $conn); header('Cache-Control: no-cache private'); header('Content-Description: File Transfer'); header('Content-Disposition: attachment, filename=climate-report.pdf'); header('Content-Type: application/pdf'); header('Content-Transfer-Encoding: binary'); header('Content-Length: ' . strlen( $result ) ); $path = realpath( "." ) . "/output.pdf"; $em = java('net.sf.jasperreports.engine.JasperExportManager'); $result = $em->exportReportToPdfFile($pm,$path); readfile( $path ); $conn->close(); Non-working Code To remove the slight redundancy (i.e., write directly to the browser), the following code looks like it should work, but it does not: $em = java('net.sf.jasperreports.engine.JasperExportManager'); $result = $em->exportReportToPdf($pm); header('Content-Length: ' . strlen( $result ) ); echo $result; Content is sent to the browser, but the file is corrupt (it begins with the PDF header) and cannot be read by any PDF reader. Question How can I take out the middle step of writing to the file and write directly to the browser so that the PDF is not corrupted? Thank you!

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  • reading binary datafile and writing into decimal no file

    - by swaroop b banerjee
    exp data is generated by my mc scaler card as a binary file with first 511 bytes as header and then 24 bit data followed by four bit roi data. i am not a expert in programming. i do understand a little. I would like to convert this file into a file (without header) decimal nos with first col as channel no (1 to 8191) then the data (24 bit) then the Roi data (4 bit). I am looking for source code in c or qbasic. thanks

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  • Free-as-in-beer binary file format inspector

    - by fbrereto
    I am looking for a utility that gives me the ability to specify a binary file format and then interpret a file of bytes according to that format. (Something along the lines of the 010 Editor, but infinitely more cost-effective). Something that runs on Mac OS X would be preferred, but I'm interested to see what all is out there in general (while more of a hassle I'd be willing to run a tool on Windows if it were superior.) What's your preference?

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  • 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.

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  • Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

    - by Justin L.
    It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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  • Binary to Date (C#) 64 Bit Format

    - by Veskechky
    We have a binary file from which we have identified the following dates (as Int64). We now the following facts about the Date/Time format; The 64 bit Date has a resolution to the microsecond The 64 bit Date has a range of 4095 years The Int64 9053167636875050944 (0x7DA34FFFFFFFFFC0) = 9th March 2010 The Int64 9053176432968073152 (0x7DA357FFFFFFFFC0) = 10th March 2010 The Int64 9053185229061095360 (0x7DA35FFFFFFFFFC0) = 11th March 2010 The Int64 9053194025154117568 (0x7DA367FFFFFFFFC0) = 12th March 2010 Any help on figuring out a way to convert this to a valid C# Date/Time is appreciated.

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  • Convert 64bit Binary to Long equivalent

    - by washtik
    How can we convert the following 64 bit binary into the long equivalent; 01111101 10100011 01001111 11111111 11111111 11111111 11111111 11000000 equals 7D A3 4F FF FF FF FF C0 HEX equals 9053167636875050944 << this is the value we want in a C# variable

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