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  • What is operator<< <> in C++?

    - by Austin Hyde
    I have seen this in a few places, and to confirm I wasn't crazy, I looked for other examples. Apparently this can come in other flavors as well, eg operator+ <>. However, nothing I have seen anywhere mentions what it is, so I thought I'd ask. It's not the easiest thing to google operator<< <>( :-)

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  • C++ segmentation error when first parameter is null in comparison operator overload

    - by user1774515
    I am writing a class called Word, that handles a c string and overloads the <, , <=, = operators. word.h: friend bool operator<(const Word &a, const Word &b); word.cc: bool operator<(const Word &a, const Word &b) { if(a == NULL && b == NULL) return false; if(a == NULL) return true; if(b == NULL) return false; return a.wd < b.wd; //wd is a valid c string } main: char* temp = NULL; //EDIT: i was mistaken, temp is a char pointer Word a("blah"); //a.wd = [b,l,a,h] cout << (temp<a); i get a segmentation error before the first line of the operator< method after the last line in the main. I can correct the problem by writing cout << (a>temp); where the operator> is similarly defined and i get no errors. but my assignment requires (temp < a) to work so this is where i ask for help. EDIT: i made a mistake the first time and i said temp was of type Word, but it is actually of type char*. so i assume that the compiler converts temp to a Word using one of my constructors. i dont know which one it would use and why this would work since the first parameter is not Word. here is the constructor i think is being used to make the Word using temp: Word::Word(char* c, char* delimeters=NULL) { char *temporary = "\0"; if(c == NULL) c = temporary; check(stoppers!=NULL, "(Word(char*,char*))NULL pointer"); //exits the program if the expression is false if(strlen(c) == 0) size = DEFAULT_SIZE; //10 else size = strlen(c) + 1 + DEFAULT_SIZE; wd = new char[size]; check(wd!=NULL, "Word(char*,char*))heap overflow"); delimiters = new char[strlen(stoppers) + 1]; //EDIT: changed to [] check(delimiters!=NULL,"Word(char*,char*))heap overflow"); strcpy(wd,c); strcpy(delimiters,stoppers); count = strlen(wd); } wd is of type char* thanks for looking at this big question and trying to help. let me know if you need more code to look at

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  • Why can operator-> be overloaded manually?

    - by FredOverflow
    Wouldn't it make sense if p->m was just syntactic sugar for (*p).m? Essentially, every operator-> that I have ever written could have been implemented as follows: Foo::Foo* operator->() { return &**this; } Is there any case where I would want p->m to mean something else than (*p).m?

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  • How to push_back without operator=() for const members?

    - by WilliamKF
    How to push_back() to a C++ std::vector without using operator=() for which the default definition violates having const members? struct Item { Item(int value) : _value(value) { } const int _value; } vector<Item> items; items.push_back(Item(3)); I'd like to keep the _value const since it should not change after the object is constructed, so the question is how do I initialize my vector with elements without invoking operator=()?

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  • UnboundLocalError: local variable 'rows' referenced before assignment

    - by patrick
    i'm trying to make a database connection by an other script. But the script didn't work propperly. and if I do a 'print' on the rows then I get the value 'null' But if I use a 'select * from incidents' query then i get the result from the table incidents. import database rows = database.database("INSERT INTO incidents VALUES(3 ,'test_title1', 'test', TO_DATE('25-07-2012', 'DD-MM-YYYY'), CURRENT_TIMESTAMP, 'sector', 50, 60)") #print database.database() print rows database.py script: import psycopg2 import sys import logfile def database(query): logfile.log(20, 'database.py', 'Executing...') con = None try: con = psycopg2.connect(database='incidents', user='ipfit5', password='tester') cur = con.cursor() #print query cur.execute(query) rows = cur.fetchall() con.commit() #test row does work #cur.execute("INSERT INTO incidents VALUES(3 ,'test_titel1', 'test', TO_DATE('25-07-2012', 'DD-MM-YYYY'), CURRENT_TIMESTAMP, 'sector', 50, 60)") except: logfile.log(40, 'database.py', 'Er is iets mis gegaan') logfile.log(40, 'database.py', str(sys.exc_info())) finally: if con: con.close() return rows

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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  • Polynomial division overloading operator

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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  • memcpy vs assignment in C

    - by SetJmp
    Under what circumstances should I expect memcpys to outperform assignments on modern INTEL/AMD hardware? I am using GCC 4.2.x on a 32 bit Intel platform (but am interested in 64 bit as well).

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  • Variable Assignment and loops (Java)

    - by Raven Dreamer
    Greetings Stack Overflowers, A while back, I was working on a program that hashed values into a hashtable (I don't remember the specifics, and the specifics themselves are irrelevant to the question at hand). Anyway, I had the following code as part of a "recordInput" method. tempElement = new hashElement(someInt); while(in.hasNext() == true) { int firstVal = in.nextInt(); if (firstVal == -911) { break; } tempElement.setKeyValue(firstVal, 0); for(int i = 1; i<numKeyValues;i++) { tempElement.setKeyValue(in.nextInt(), i); } elementArray[placeValue] = tempElement; placeValue++; } // close while loop } // close method This part of the code was giving me a very nasty bug -- no matter how I finagled it, no matter what input I gave the program, it would always produce an array full of only a single value -- the last one. The problem, as I later determined it, was that because I had not created the tempElement variable within the loop, and because values were not being assigned to elementArray[] until after the loop had ended -- every term was defined rather as "tempElement" -- when the loop terminated, every slot in the array was filled with the last value tempElement had taken. I was able to fix this bug by moving the declaration of tempElement within the while loop. My question to you, Stackoverflow, is whether there is another (read: better) way to avoid this bug while keeping the variable declaration of tempElement outside the while loop. (suggestions for better title and tags also appreciated)

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  • Java assignment issues - Is this atomic?

    - by Bob
    Hi, I've got some questions about Java's assigment. Strings I've got a class: public class Test { private String s; public synchronized void setS(String str){ s = s + " - " + str; } public String getS(){ return s; } } I'm using "synchronized" in my setter, and avoiding it in my getter, because in my app, there are a tons of data gettings, and very few settings. Settings must be synchronized to avoid inconsistency. My question is: is getting and setting a variable atomic? I mean, in a multithreaded environment, Thread1 is about to set variable s, while Thread2 is about to get "s". Is there any way the getter method could get something different than the s's old value or the s's new value (suppose we've got only two threads)? In my app it is not a problem to get the new value, and it is not a problem to get the old one. But could I get something else? What about HashMap's getting and putting? considering this: public class Test { private Map<Integer, String> map = Collections.synchronizedMap(new HashMap<Integer, String>()); public synchronized void setMapElement(Integer key, String value){ map.put(key, value); } public String getValue(Integer key){ return map.get(key); } } Is putting and getting atomic? How does HashMap handle putting an element into it? Does it first remove the old value and put the now one? Could I get other than the old value or the new value? Thanks in advance!

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  • Java array assignment (multiple values)

    - by Danny King
    Hello, I have a Java array defined already e.g. float[] values = new float[3]; I would like to do something like this further on in the code: values = {0.1f, 0.2f, 0.3f}; But that gives me a compile error. Is there a nicer way to define multiple values at once, rather than doing this?: values[0] = 0.1f; values[1] = 0.2f; values[2] = 0.3f; Thanks!

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  • How do I do multiple assignment in MATLAB?

    - by Benjamin Oakes
    Here's an example of what I'm looking for: >> foo = [88, 12]; >> [x, y] = foo; I'd expect something like this afterwards: >> x x = 88 >> y y = 12 But instead I get errors like: ??? Too many output arguments. I thought deal() might do it, but it seems to only work on cells. >> [x, y] = deal(foo{:}); ??? Cell contents reference from a non-cell array object. How do I solve my problem? Must I constantly index by 1 and 2 if I want to deal with them separately?

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  • Bizzare Java invalid Assignment Operator Error

    - by Kay
    public class MaxHeap<T extends Comparable<T>> implements Heap<T>{ private T[] heap; private int lastIndex; private static final int defaultInitialCapacity = 25; public void add(T newItem) throws HeapException{ if (lastIndex < Max_Heap){ heap[lastIndex] = newItem; int place = lastIndex; int parent = (place – 1)/2; //ERROR HERE********** while ( (parent >=0) && (heap[place].compareTo(heap[parent])>0)){ T temp = heap[place]; heap[place] = heap[parent]; heap[parent] = temp; place = parent; parent = (place-1)/2; }else { throw new HeapException(“HeapException: Heap full”); } } } Eclipse complains that there is a: "Syntax error on token "Invalid Character", invalid AssignmentOperator" With the red line beneath the '(place-1)' There shouldn't be an error at all since it's just straight-forward arithmetic. Or is it not that simple?

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  • C++ assignment - stylish or performance?

    - by joejax
    Having been writing Java code for many years, I was amazed when I see this C++ statement: int a,b; int c = (a=1, b=a+2, b*3); My question is: Is this a choice of coding style, or it has real benefit? (looking for a practicle use case) I think the compiler will see it the same as following: int a=1, b=a+2; int c = b*3; (What's the offical name for this? I assume it's a standard C/C++ syntax.)

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  • Java: Object Array assignment in for loop

    - by Hackster
    I am trying to use Dijkstra's algorithm to find the shortest path from a specific vertex (v0) to the rest of them. That is solved and works well with this code from this link below: http://en.literateprograms.org/index.php?title=Special:DownloadCode/Dijkstra%27s_algorithm_(Java)&oldid=15444 I am having trouble with assigning the Edge array in a for loop from the user input, as opposed to hard-coding it like it is here. Any help assigning a new edge to Edge[] adjacencies from each vertex? Keeping in mind it could be 1 or multiple edges. class Vertex implements Comparable<Vertex> { public final String name; public Edge[] adjacencies; public double minDistance = Double.POSITIVE_INFINITY; public Vertex previous; public Vertex(String argName) { name = argName; } public String toString() { return name; } public int compareTo(Vertex other){ return Double.compare(minDistance, other.minDistance); } } class Edge{ public final Vertex target; public final double weight; public Edge(Vertex argTarget, double argWeight){ target = argTarget; weight = argWeight; } } public static void main(String[] args) { Vertex v[] = new Vertex[3]; Vertex v[0] = new Vertex("Harrisburg"); Vertex v[1] = new Vertex("Baltimore"); Vertex v[2] = new Vertex("Washington"); v0.adjacencies = new Edge[]{ new Edge(v[1], 1), new Edge(v[2], 3) }; v1.adjacencies = new Edge[]{ new Edge(v[0], 1), new Edge(v[2], 1),}; v2.adjacencies = new Edge[]{ new Edge(v[0], 3), new Edge(v[1], 1) }; Vertex[] vertices = { v0, v1, v2}; /*Three vertices with weight: V0 connects (V1,1),(V2,3) V1 connects (V0,1),(V2,1) V2 connects (V1,1),(V2,3) */ computePaths(v0); for (Vertex v : vertices){ System.out.println("Distance to " + v + ": " + v.minDistance); List<Vertex> path = getShortestPathTo(v); System.out.println("Path: " + path); } } } The above code works well in finding the shortest path from v0 to all the other vertices. The problem occurs when assigning the new edge[] to edge[] adjacencies. For example this does not produce the correct output: for (int i = 0; i < total_vertices; i++){ s = br.readLine(); char[] line = s.toCharArray(); for (int j = 0; j < line.length; j++){ if(j % 4 == 0 ){ //Input: vertex weight vertex weight: 1 1 2 3 int vert = Integer.parseInt(String.valueOf(line[j])); int w = Integer.parseInt(String.valueOf(line[j+2])); v[i].adjacencies = new Edge[] {new Edge(v[vert], w)}; } } } As opposed to this: v0.adjacencies = new Edge[]{ new Edge(v[1], 1), new Edge(v[2], 3) }; How can I take the user input and make an Edge[], to pass it to adjacencies? The problem is it could be 0 edges or many. Any help would be much appreciated Thanks!

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  • 3x3 Sobel operator and gradient features

    - by pithyless
    Reading a paper, I'm having difficulty understanding the algorithm described: Given a black and white digital image of a handwriting sample, cut out a single character to analyze. Since this can be any size, the algorithm needs to take this into account (if it will be easier, we can assume the size is 2^n x 2^m). Now, the description states given this image we will convert it to a 512-bit feature (a 512-bit hash) as follows: (192 bits) computes the gradient of the image by convolving it with a 3x3 Sobel operator. The direction of the gradient at every edge is quantized to 12 directions. (192 bits) The structural feature generator takes the gradient map and looks in a neighborhood for certain combinations of gradient values. (used to compute 8 distinct features that represent lines and corners in the image) (128 bits) Concavity generator uses an 8-point star operator to find coarse concavities in 4 directions, holes, and lagrge-scale strokes. The image feature maps are normalized with a 4x4 grid. I'm for now struggling with how to take an arbitrary image, split into 16 sections, and using a 3x3 Sobel operator to come up with 12 bits for each section. (But if you have some insight into the other parts, feel free to comment :)

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  • operator<< overload,

    - by mr.low
    //using namespace std; using std::ifstream; using std::ofstream; using std::cout; class Dog { friend ostream& operator<< (ostream&, const Dog&); public: char* name; char* breed; char* gender; Dog(); ~Dog(); }; im trying to overload the << operator. I'm also trying to practice good coding. But my code wont compile unless i uncomment the using namespace std. i keep getting this error and i dont know. im using g++ compiler. Dog.h:20: error: ISO C++ forbids declaration of ‘ostream’ with no type Dog.h:20: error: ‘ostream’ is neither function nor member function; cannot be declared friend. if i add line using std::cout; then i get this error. Dog.h:21: error: ISO C++ forbids declaration of ‘ostream’ with no type. Can somebody tell me the correct way to overload the << operator with out using namespace std;

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  • How to overload operator<< for qDebug

    - by iyo
    Hi, I'm trying to create more useful debug messages for my class where store data. My code is looking something like this #include <QAbstractTableModel> #include <QDebug> /** * Model for storing data. */ class DataModel : public QAbstractTableModel { // for debugging purposes friend QDebug & operator<< (const QDebug &d, DataModel model); //other stuff }; /** * Overloading operator for debugging purposes */ QDebug & operator<< (QDebug &d, DataModel model) { d << "Hello world!"; return d; } I expect qDebug() << model will print "Hello world!". However, there is alway something like "QAbstractTableModel(0x1c7e520)" on the output. Do you have any idea what's wrong?

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  • Providing less than operator for one element of a pair

    - by Koszalek Opalek
    What would be the most elegant way too fix the following code: #include <vector> #include <map> #include <set> using namespace std; typedef map< int, int > row_t; typedef vector< row_t > board_t; typedef row_t::iterator area_t; bool operator< ( area_t const& a, area_t const& b ) { return( a->first < b->first ); }; int main( int argc, char* argv[] ) { int row_num; area_t it; set< pair< int, area_t > > queue; queue.insert( make_pair( row_num, it ) ); // does not compile }; One way to fix it is moving the definition of less< to namespace std (I know, you are not supposed to do it.) namespace std { bool operator< ( area_t const& a, area_t const& b ) { return( a->first < b->first ); }; }; Another obvious solution is defining less than< for pair< int, area_t but I'd like to avoid that and be able to define the operator only for the one element of the pair where it is not defined.

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  • How operator oveloading works

    - by Rasmi Ranjan Nayak
    I have below code class rectangle { ..... .....//Some code int operator+(rectangle r1) { return(r1.length+length); } }; In main fun. int main() { rectangle r1(10,20); rectangle r2(40,60); rectangle r3(30,60); int len = r1+r3; } Here if we will see in operator+(), we are doing r1.length + length. How the compiler comes to know that the 2nd length in return statement belong to object r3 not to r1 or r2? I think answer may be in main() we have writeen int len = r1+r3; If that is the case then why do we need to write in operator+(....) { r1.lenth + lenth; //Why not length + length? } Why not length + length? Bcause compiler already knows from main() that the first length belong to object r1 and 2nd to object r3.

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