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  • operator overloading

    - by cpp_Beginner
    Hi, Could anybody tell me the difference between operator overloading using the friend keyword and as a member function inside a class? also what is the difference incase of any unary operator overloading i.e., as a friend and as a member function

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  • Negate the null-coalescing operator

    - by jhunter
    I have a bunch of strings I need to use .Trim() on, but they can be null. It would be much more concise if I could do something like: string endString = startString !?? startString.Trim(); Basically return the part on the right if the part on the left is NOT null, otherwise just return the null value. I just ended up using the ternary operator, but is there anyway to use the null-coalescing operator for this purpose?

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  • behaviour of the implicit copy constructor / assignment operator

    - by Tobias Langner
    Hello, I have a question regarding the C++ Standard. Suppose you have a base class with user defined copy constructor and assignment operator. The derived class uses the implicit one generated by the compiler. Does copying / assignment of the derived class call the user defined copy constructor / assignment operator? Or do you need to implement user defined versions that call the base class? Thank you for your help.

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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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  • 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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  • 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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  • 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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  • 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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  • perl - universal operator overload

    - by Todd Freed
    I have an idea for perl, and I'm trying to figure out the best way to implement it. The idea is to have new versions of every operator which consider the undefined value as the identity of that operation. For example: $a = undef + 5; # undef treated as 0, so $a = 5 $a = undef . "foo"; # undef treated as '', so $a = foo $a = undef && 1; # undef treated as false, $a = true and so forth. ideally, this would be in the language as a pragma, or something. use operators::awesome; However, I would be satisfied if I could implement this special logic myself, and then invoke it where needed: use My::Operators; The problem is that if I say "use overload" inside My::Operators only affects objects blessed into My::Operators. So the question is: is there a way (with "use overoad" or otherwise) to do a "universal operator overload" - which would be called for all operations, not just operations on blessed scalars. If not - who thinks this would be a great idea !? It would save me a TON of this kind of code if($object && $object{value} && $object{value} == 15) replace with if($object{value} == 15) ## the special "is-equal-to" operator

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  • Recursion problem overloading an operator

    - by Tronfi
    I have this: typedef string domanin_name; And then, I try to overload the operator< in this way: bool operator<(const domain_name & left, const domain_name & right){ int pos_label_left = left.find_last_of('.'); int pos_label_right = right.find_last_of('.'); string label_left = left.substr(pos_label_left); string label_right = right.substr(pos_label_right); int last_pos_label_left=0, last_pos_label_right=0; while(pos_label_left!=string::npos && pos_label_right!=string::npos){ if(label_left<label_right) return true; else if(label_left>label_right) return false; else{ last_pos_label_left = pos_label_left; last_pos_label_right = pos_label_right; pos_label_left = left.find_last_of('.', last_pos_label_left); pos_label_right = right.find_last_of('.', last_pos_label_left); label_left = left.substr(pos_label_left, last_pos_label_left); label_right = right.substr(pos_label_right, last_pos_label_right); } } } I know it's a strange way to overload the operator <, but I have to do it this way. It should do what I want. That's not the point. The problem is that it enter in an infinite loop right in this line: if(label_left<label_right) return true; It seems like it's trying to use this overloading function itself to do the comparision, but label_left is a string, not a domain name! Any suggestion?

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  • overload == (and != , of course) operator, can I bypass == to determine whether the object is null

    - by LLS
    Hello, when I try to overload operator == and != in C#, and override Equal as recommended, I found I have no way to distinguish a normal object and null. For example, I defined a class Complex. public static bool operator ==(Complex lhs, Complex rhs) { return lhs.Equals(rhs); } public static bool operator !=(Complex lhs, Complex rhs) { return !lhs.Equals(rhs); } public override bool Equals(object obj) { if (obj is Complex) { return (((Complex)obj).Real == this.Real && ((Complex)obj).Imaginary == this.Imaginary); } else { return false; } } But when I want to use if (temp == null) When temp is really null, some exception happens. And I can't use == to determine whether the lhs is null, which will cause infinite loop. What should I do in this situation. One way I can think of is to us some thing like Class.Equal(object, object) (if it exists) to bypass the == when I do the check. What is the normal way to solve the problem?

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  • Operator Overloading << in c++

    - by thlgood
    I'm a fresh man in C++. I write this simple program to practice Overlaoding. This is my code: #include <iostream> #include <string> using namespace std; class sex_t { private: char __sex__; public: sex_t(char sex_v = 'M'):__sex__(sex_v) { if (sex_v != 'M' && sex_v != 'F') { cerr << "Sex type error!" << sex_v << endl; __sex__ = 'M'; } } const ostream& operator << (const ostream& stream) { if (__sex__ == 'M') cout << "Male"; else cout << "Female"; return stream; } }; int main(int argc, char *argv[]) { sex_t me('M'); cout << me << endl; return 0; } When I compiler it, It print a lots of error message: The error message was in a mess. It's too hard for me to found useful message sex.cpp: ???‘int main(int, char**)’?: sex.cpp:32:10: ??: ‘operator<<’?‘std::cout << me’????? sex.cpp:32:10: ??: ???: /usr/include/c++/4.6/ostream:110:7: ??: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostre

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  • Layout Columns - Equal Height

    - by Kyle
    I remember first starting out using tables for layouts and learned that I should not be doing that. I am working on a new site and can not seem to do equal height columns without using tables. Here is an example of the attempt with div tags. <div class="row"> <div class="column">column1</div> <div class="column">column2</div> <div class="column">column3</div> <div style="clear:both"></div> </div> Now what I tried with that was doing making columns float left and setting their widths to 33% which works fine, I use the clear:both div so that the row would be the size of the biggest column, but the columns will be different sizes based on how much content they have. I have found many fixes which mostly involve css hacks and just making it look like its right but that's not what I want. I thought of just doing it in javascript but then it would look different for those who choose to disable their javascript. The only true way of doing it that I can think of is using tables since the cells all have equal heights in the same row. But I know its bad to use tables. After searching forever I than came across this: http://intangiblestyle.com/lab/equal-height-columns-with-css/ What it seems to do is exactly the same as tables since its just setting its display exactly like tables. Would using that be just as bad as using tables? I honestly can't find anything else that I could do. edit @Su' I have looked into "faux columns" and do not think that is what I want. I think I would be able to implement better designs for my site using the display:table method. I posted this question because I just wasn't sure if I should since I have always heard its bad using tables in website layouts.

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  • Class Assignment Operators

    - by Maxpm
    I made the following operator overloading test: #include <iostream> #include <string> using namespace std; class TestClass { string ClassName; public: TestClass(string Name) { ClassName = Name; cout << ClassName << " constructed." << endl; } ~TestClass() { cout << ClassName << " destructed." << endl; } void operator=(TestClass Other) { cout << ClassName << " in operator=" << endl; cout << "The address of the other class is " << &Other << "." << endl; } }; int main() { TestClass FirstInstance("FirstInstance"); TestClass SecondInstance("SecondInstance"); FirstInstance = SecondInstance; SecondInstance = FirstInstance; return 0; } The assignment operator behaves as-expected, outputting the address of the other class. Now, how would I actually assign something from the other class? For example, something like this: void operator=(TestClass Other) { ClassName = Other.ClassName; }

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  • Why overload true and false instead of defining bool operator?

    - by Joe Enos
    I've been reading about overloading true and false in C#, and I think I understand the basic difference between this and defining a bool operator. The example I see around is something like: public static bool operator true(Foo foo) { return (foo.PropA > 0); } public static bool operator false(Foo foo) { return (foo.PropA <= 0); } To me, this is the same as saying: public static implicit operator bool(Foo foo) { return (foo.PropA > 0); } The difference, as far as I can tell, is that by defining true and false separately, you can have an object that is both true and false, or neither true nor false: public static bool operator true(Foo foo) { return true; } public static bool operator false(Foo foo) { return true; } //or public static bool operator true(Foo foo) { return false; } public static bool operator false(Foo foo) { return false; } I'm sure there's a reason this is allowed, but I just can't think of what it is. To me, if you want an object to be able to be converted to true or false, a single bool operator makes the most sense. Can anyone give me a scenario where it makes sense to do it the other way? Thanks

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