Search Results

Search found 2479 results on 100 pages for 'the operator'.

Page 2/100 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

• C++ Operator Ambiguity

  - by Scott

  Forgive me, for I am fairly new to C++, but I am having some trouble regarding operator ambiguity. I think it is compiler-specific, for the code compiled on my desktop. However, it fails to compile on my laptop. I think I know what's going wrong, but I don't see an elegant way around it. Please let me know if I am making an obvious mistake. Anyhow, here's what I'm trying to do: I have made my own vector class called Vector4 which looks something like this: class Vector4 { private: GLfloat vector[4]; ... } Then I have these operators, which are causing the problem: operator GLfloat* () { return vector; } operator const GLfloat* () const { return vector; } GLfloat& operator [] (const size_t i) { return vector[i]; } const GLfloat& operator [] (const size_t i) const { return vector[i]; } I have the conversion operator so that I can pass an instance of my Vector4 class to glVertex3fv, and I have subscripting for obvious reasons. However, calls that involve subscripting the Vector4 become ambiguous to the compiler: enum {x, y, z, w} Vector4 v(1.0, 2.0, 3.0, 4.0); glTranslatef(v[x], v[y], v[z]); Here are the candidates: candidate 1: const GLfloat& Vector4:: operator[](size_t) const candidate 2: operator[](const GLfloat*, int) <built-in> Why would it try to convert my Vector4 to a GLfloat* first when the subscript operator is already defined on Vector4? Is there a simple way around this that doesn't involve typecasting? Am I just making a silly mistake? Thanks for any help in advance.

  Read the article

• wrong operator() overload called

  - by user313202

  okay, I am writing a matrix class and have overloaded the function call operator twice. The core of the matrix is a 2D double array. I am using the MinGW GCC compiler called from a windows console. the first overload is meant to return a double from the array (for viewing an element). the second overload is meant to return a reference to a location in the array (for changing the data in that location. double operator()(int row, int col) const ; //allows view of element double &operator()(int row, int col); //allows assignment of element I am writing a testing routine and have discovered that the "viewing" overload never gets called. for some reason the compiler "defaults" to calling the overload that returns a reference when the following printf() statement is used. fprintf(outp, "%6.2f\t", testMatD(i,j)); I understand that I'm insulting the gods by writing my own matrix class without using vectors and testing with C I/O functions. I will be punished thoroughly in the afterlife, no need to do it here. Ultimately I'd like to know what is going on here and how to fix it. I'd prefer to use the cleaner looking operator overloads rather than member functions. Any ideas? -Cal the matrix class: irrelevant code omitted class Matrix { public: double getElement(int row, int col)const; //returns the element at row,col //operator overloads double operator()(int row, int col) const ; //allows view of element double &operator()(int row, int col); //allows assignment of element private: //data members double **array; //pointer to data array }; double Matrix::getElement(int row, int col)const{ //transform indices into true coordinates (from sorted coordinates //only row needs to be transformed (user can only sort by row) row = sortedArray[row]; result = array[usrZeroRow+row][usrZeroCol+col]; return result; } //operator overloads double Matrix::operator()(int row, int col) const { //this overload is used when viewing an element return getElement(row,col); } double &Matrix::operator()(int row, int col){ //this overload is used when placing an element return array[row+usrZeroRow][col+usrZeroCol]; } The testing program: irrelevant code omitted int main(void){ FILE *outp; outp = fopen("test_output.txt", "w+"); Matrix testMatD(5,7); //construct 5x7 matrix //some initializations omitted fprintf(outp, "%6.2f\t", testMatD(i,j)); //calls the wrong overload }

  Read the article

• Operator Overloading in C

  - by Leif Andersen

  In C++, I can change the operator on a specific class by doing something like this: MyClass::operator==/*Or some other operator such as =, >, etc.*/(Const MyClass rhs) { /* Do Stuff*/; } But with there being no classes (built in by default) in C. So, how could I do operator overloading for just general functions? For example, if I remember correctly, importing stdlib.h gives you the - operator, which is just syntactic sugar for (*strcut_name).struct_element. So how can I do this in C? Thank you.

  Read the article

• Explain this C++ operator definition

  - by David Johnstone

  I have the following operator defined in a C++ class called StringProxy: operator std::string&() { return m_string; } a) What is this and how does this work? I understand the idea of operator overloading, but they normally look like X operator+(double i). b) Given an instance of StringProxy, how can I use this operator to get the m_string?

  Read the article

• C++ Operator overloading - 'recreating the Vector'

  - by Wallter

  I am currently in a collage second level programing course... We are working on operator overloading... to do this we are to rebuild the vector class... I was building the class and found that most of it is based on the [] operator. When I was trying to implement the + operator I run into a weird error that my professor has not seen before (apparently since the class switched IDE's from MinGW to VS express...) (I am using Visual Studio Express 2008 C++ edition...) Vector.h #include <string> #include <iostream> using namespace std; #ifndef _VECTOR_H #define _VECTOR_H const int DEFAULT_VECTOR_SIZE = 5; class Vector { private: int * data; int size; int comp; public: inline Vector (int Comp = 5,int Size = 0) : comp(Comp), size(Size) { if (comp > 0) { data = new int [comp]; } else { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; } } int size_ () const { return size; } int comp_ () const { return comp; } bool push_back (int); bool push_front (int); void expand (); void expand (int); void clear (); const string at (int); int operator[ ](int); Vector& operator+ (Vector&); Vector& operator- (const Vector&); bool operator== (const Vector&); bool operator!= (const Vector&); ~Vector() { delete [] data; } }; ostream& operator<< (ostream&, const Vector&); #endif Vector.cpp #include <iostream> #include <string> #include "Vector.h" using namespace std; const string Vector::at(int i) { this[i]; } void Vector::expand() { expand(size); } void Vector::expand(int n ) { int * newdata = new int [comp * 2]; if (*data != NULL) { for (int i = 0; i <= (comp); i++) { newdata[i] = data[i]; } newdata -= comp; comp += n; delete [] data; *data = *newdata; } else if ( *data == NULL || comp == 0) { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; size = 0; } } bool Vector::push_back(int n) { if (comp = 0) { expand(); } for (int k = 0; k != 2; k++) { if ( size != comp ){ data[size] = n; size++; return true; } else { expand(); } } return false; } void Vector::clear() { delete [] data; comp = 0; size = 0; } int Vector::operator[] (int place) { return (data[place]); } Vector& Vector::operator+ (Vector& n) { int temp_int = 0; if (size > n.size_() || size == n.size_()) { temp_int = size; } else if (size < n.size_()) { temp_int = n.size_(); } Vector newone(temp_int); int temp_2_int = 0; for ( int j = 0; j <= temp_int && j <= n.size_() && j <= size; j++) { temp_2_int = n[j] + data[j]; newone[j] = temp_2_int; } //////////////////////////////////////////////////////////// return newone; //////////////////////////////////////////////////////////// } ostream& operator<< (ostream& out, const Vector& n) { for (int i = 0; i <= n.size_(); i++) { //////////////////////////////////////////////////////////// out << n[i] << " "; //////////////////////////////////////////////////////////// } return out; } Errors: out << n[i] << " "; error C2678: binary '[' : no operator found which takes a left-hand operand of type 'const Vector' (or there is no acceptable conversion) return newone; error C2106: '=' : left operand must be l-value As stated above, I am a student going into Computer Science as my selected major I would appreciate tips, pointers, and better ways to do stuff :D

  Read the article

• Are free operator->* overloads evil?

  - by Potatoswatter

  I was perusing section 13.5 after refuting the notion that built-in operators do not participate in overload resolution, and noticed that there is no section on operator->*. It is just a generic binary operator. Its brethren, operator->, operator*, and operator[], are all required to be non-static member functions. This precludes definition of a free function overload to an operator commonly used to obtain a reference from an object. But the uncommon operator->* is left out. In particular, operator[] has many similarities. It is binary (they missed a golden opportunity to make it n-ary), and it accepts some kind of container on the left and some kind of locator on the right. Its special-rules section, 13.5.5, doesn't seem to have any actual effect except to outlaw free functions. (And that restriction even precludes support for commutativity!) So, for example, this is perfectly legal (in C++0x, remove obvious stuff to translate to C++03): #include <utility> #include <iostream> #include <type_traits> using namespace std; template< class F, class S > typename common_type< F,S >::type operator->*( pair<F,S> const &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( pair<T,T> &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( bool l, pair<T,T> &r ) { return l? r.second : r.first; } int main() { auto x = make_pair( 1, 2.3 ); cerr << x->*false << " " << x->*4 << endl; auto y = make_pair( 5, 6 ); y->*(0) = 7; y->*0->*y = 8; // evaluates to 7->*y = y.second cerr << y.first << " " << y.second << endl; } I can certainly imagine myself giving into temp[la]tation. For example, scaled indexes for vector: v->*matrix_width[2][5] = x; Did the standards committee forget to prevent this, was it considered too ugly to bother, or are there real-world use cases?

  Read the article

• Overloading *(iterator + n) and *(n + iterator) in a C++ iterator class?

  - by exscape

  (Note: I'm writing this project for learning only; comments about it being redundant are... uh, redundant. ;) I'm trying to implement a random access iterator, but I've found very little literature on the subject, so I'm going by trial and error combined with Wikpedias list of operator overload prototypes. It's worked well enough so far, but I've hit a snag. Code such as exscape::string::iterator i = string_instance.begin(); std::cout << *i << std::endl; works, and prints the first character of the string. However, *(i + 1) doesn't work, and neither does *(1 + i). My full implementation would obviously be a bit too much, but here's the gist of it: namespace exscape { class string { friend class iterator; ... public: class iterator : public std::iterator<std::random_access_iterator_tag, char> { ... char &operator*(void) { return *p; // After some bounds checking } char *operator->(void) { return p; } char &operator[](const int offset) { return *(p + offset); // After some bounds checking } iterator &operator+=(const int offset) { p += offset; return *this; } const iterator operator+(const int offset) { iterator out (*this); out += offset; return out; } }; }; } int main() { exscape::string s = "ABCDEF"; exscape::string::iterator i = s.begin(); std::cout << *(i + 2) << std::endl; } The above fails with (line 632 is, of course, the *(i + 2) line): string.cpp: In function ‘int main()’: string.cpp:632: error: no match for ‘operator*’ in ‘*exscape::string::iterator::operator+(int)(2)’ string.cpp:105: note: candidates are: char& exscape::string::iterator::operator*() *(2 + i) fails with: string.cpp: In function ‘int main()’: string.cpp:632: error: no match for ‘operator+’ in ‘2 + i’ string.cpp:434: note: candidates are: exscape::string exscape::operator+(const char*, const exscape::string&) My guess is that I need to do some more overloading, but I'm not sure what operator I'm missing.

  Read the article

• How do you override operator == when using interfaces instead of actual types?

  - by RickL

  I have some code like this: How should I implement the operator == so that it will be called when the variables are of interface IMyClass? public class MyClass : IMyClass { public static bool operator ==(MyClass a, MyClass b) { if (ReferenceEquals(a, b)) return true; if ((Object)a == null || (Object)b == null) return false; return false; } public static bool operator !=(MyClass a, MyClass b) { return !(a == b); } } class Program { static void Main(string[] args) { IMyClass m1 = new MyClass(); IMyClass m2 = new MyClass(); MyClass m3 = new MyClass(); MyClass m4 = new MyClass(); Console.WriteLine(m1 == m2); // does not go into custom == function. why not? Console.WriteLine(m3 == m4); // DOES go into custom == function } }

  Read the article

• friending istream operator with class

  - by user1388172

  hello i'm trying to overload my operator >> to my class but i ecnouter an error in eclipse. code: friend istream& operator>>(const istream& is, const RAngle& ra){ return is >> ra.x >> ra.y; } code2: friend istream& operator>>(const istream& is, const RAngle& ra) { is >> ra.x; is >> ra.y; return is } Both crash and i don't know why, please help. EDIT: ra.x & ra.y are both 2 private ints of my class; Full error: error: ..\/rightangle.h: In function 'std::istream& operator>>(std::istream&, const RAngle&)': ..\/rightangle.h:65:12: error: ambiguous overload for 'operator>>' in 'is >> ra.RAngle::x' ..\/rightangle.h:65:12: note: candidates are: c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:122:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__istream_type& (*)(std::basic_istream<_CharT, _Traits>::__istream_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:122:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__istream_type& (*)(std::basic_istream<char>::__istream_type&) {aka std::basic_istream<char>& (*)(std::basic_istream<char>&)}' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:126:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__ios_type& (*)(std::basic_istream<_CharT, _Traits>::__ios_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>, std::basic_istream<_CharT, _Traits>::__ios_type = std::basic_ios<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:126:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__ios_type& (*)(std::basic_istream<char>::__ios_type&) {aka std::basic_ios<char>& (*)(std::basic_ios<char>&)}' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:133:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:133:7: note: no known conversion for argument 1 from 'const int' to 'std::ios_base& (*)(std::ios_base&)' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:241:7: note: std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__streambuf_type*) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__streambuf_type = std::basic_streambuf<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:241:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__streambuf_type* {aka std::basic_streambuf<char>*}' ..\/rightangle.h:66:12: error: ambiguous overload for 'operator>>' in 'is >> ra.RAngle::y' ..\/rightangle.h:66:12: note: candidates are: c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:122:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__istream_type& (*)(std::basic_istream<_CharT, _Traits>::__istream_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:122:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__istream_type& (*)(std::basic_istream<char>::__istream_type&) {aka std::basic_istream<char>& (*)(std::basic_istream<char>&)}' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:126:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__ios_type& (*)(std::basic_istream<_CharT, _Traits>::__ios_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>, std::basic_istream<_CharT, _Traits>::__ios_type = std::basic_ios<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:126:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__ios_type& (*)(std::basic_istream<char>::__ios_type&) {aka std::basic_ios<char>& (*)(std::basic_ios<char>&)}' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:133:7: note: std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__istream_type = std::basic_istream<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:133:7: note: no known conversion for argument 1 from 'const int' to 'std::ios_base& (*)(std::ios_base&)' c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:241:7: note: std::basic_istream<_CharT, _Traits>& std::basic_istream<_CharT, _Traits>::operator>>(std::basic_istream<_CharT, _Traits>::__streambuf_type*) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_istream<_CharT, _Traits>::__streambuf_type = std::basic_streambuf<char>] <near match> c:\mingw\bin\../lib/gcc/mingw32/4.6.1/include/c++/istream:241:7: note: no known conversion for argument 1 from 'const int' to 'std::basic_istream<char>::__streambuf_type* {aka std::basic_streambuf<char>*}''

  Read the article

• User defined conversion operator as argument for printf

  - by BC

  I have a class that defined a user defined operator for a TCHAR*, like so CMyClass::operator const TCHAR*() const { // returns text as const TCHAR* } I want to be able to do something like CMyClass myClass; _tprintf(_T("%s"), myClass); or even _tprintf(_T("%s"), CMyClass(value)); But when trying, printf always prints (null) instead of the value. I have also tried a normal char* operator, as well variations with const etc. It only works correctly if I explicitly call the operator or do a cast, like _tprintf(_T("%s\n"), (const TCHAR*)myClass); _tprintf(_T("%s\n"), myClass.operator const TCHAR *()); However, I don't want to cast. How can this be achieved? Note, that a possibility is to create a function that has a parameter of const TCHAR*, so that it forcible calls the operator TCHAR*, but this I also don't want to implement.

  Read the article

• How to reduce redundant code when adding new c++0x rvalue reference operator overloads

  - by Inverse

  I am adding new operator overloads to take advantage of c++0x rvalue references, and I feel like I'm producing a lot of redundant code. I have a class, tree, that holds a tree of algebraic operations on double values. Here is an example use case: tree x = 1.23; tree y = 8.19; tree z = (x + y)/67.31 - 3.15*y; ... std::cout << z; // prints "(1.23 + 8.19)/67.31 - 3.15*8.19" For each binary operation (like plus), each side can be either an lvalue tree, rvalue tree, or double. This results in 8 overloads for each binary operation: // core rvalue overloads for plus: tree operator +(const tree& a, const tree& b); tree operator +(const tree& a, tree&& b); tree operator +(tree&& a, const tree& b); tree operator +(tree&& a, tree&& b); // cast and forward cases: tree operator +(const tree& a, double b) { return a + tree(b); } tree operator +(double a, const tree& b) { return tree(a) + b; } tree operator +(tree&& a, double b) { return std::move(a) + tree(b); } tree operator +(double a, tree&& b) { return tree(a) + std::move(b); } // 8 more overloads for minus // 8 more overloads for multiply // 8 more overloads for divide // etc which also has to be repeated in a way for each binary operation (minus, multiply, divide, etc). As you can see, there are really only 4 functions I actually need to write; the other 4 can cast and forward to the core cases. Do you have any suggestions for reducing the size of this code? PS: The class is actually more complex than just a tree of doubles. Reducing copies does dramatically improve performance of my project. So, the rvalue overloads are worthwhile for me, even with the extra code. I have a suspicion that there might be a way to template away the "cast and forward" cases above, but I can't seem to think of anything.

  Read the article

• C++ stream as a parameter when overloading operator<<

  - by TheOm3ga

  I'm trying to write my own logging class and use it as a stream: logger L; L << "whatever" << std::endl; This is the code I started with: #include <iostream> using namespace std; class logger{ public: template <typename T> friend logger& operator <<(logger& log, const T& value); }; template <typename T> logger& operator <<(logger& log, T const & value) { // Here I'd output the values to a file and stdout, etc. cout << value; return log; } int main(int argc, char *argv[]) { logger L; L << "hello" << '\n' ; // This works L << "bye" << "alo" << endl; // This doesn't work return 0; } But I was getting an error when trying to compile, saying that there was no definition for operator<<: pruebaLog.cpp:31: error: no match for ‘operator<<’ in ‘operator<< [with T = char [4]](((logger&)((logger*)operator<< [with T = char [4]](((logger&)(& L)), ((const char (&)[4])"bye")))), ((const char (&)[4])"alo")) << std::endl’ So, I've been trying to overload operator<< to accept this kind of streams, but it's driving me mad. I don't know how to do it. I've been loking at, for instance, the definition of std::endl at the ostream header file and written a function with this header: logger& operator <<(logger& log, const basic_ostream<char,char_traits<char> >& (*s)(basic_ostream<char,char_traits<char> >&)) But no luck. I've tried the same using templates instead of directly using char, and also tried simply using "const ostream& os", and nothing. Another thing that bugs me is that, in the error output, the first argument for operator<< changes, sometimes it's a reference to a pointer, sometimes looks like a double reference...

  Read the article

• whats the differnce between dot operator and scope resolution operator

  - by oDx

  I just wanted to know the difference between . operator and :: operator?

  Read the article

• C++ Unary - Operator Overload Won't Compile

  - by Brian Hooper

  I am attempting to create an overloaded unary - operator but can't get the code to compile. A cut-down version of the code is as follows:- class frag { public: frag myfunc (frag oper1, frag oper2); frag myfunc2 (frag oper1, frag oper2); friend frag operator + (frag &oper1, frag &oper2); frag operator - () { frag f; f.element = -element; return f; } private: int element; }; frag myfunc (frag oper1, frag oper2) { return oper1 + -oper2; } frag myfunc2 (frag oper1, frag oper2) { return oper1 + oper2; } frag operator+ (frag &oper1, frag &oper2) { frag innerfrag; innerfrag.element = oper1.element + oper2.element; return innerfrag; } The compiler reports... /home/brian/Desktop/frag.hpp: In function ‘frag myfunc(frag, frag)’: /home/brian/Desktop/frag.hpp:41: error: no match for ‘operator+’ in ‘oper1 + oper2.frag::operator-()’ /home/brian/Desktop/frag.hpp:16: note: candidates are: frag operator+(frag&, frag&) Could anyone suggest what I need to be doing here? Thanks.

  Read the article

• operator overloading and inheritance

  - by user168715

  I was given the following code: class FibHeapNode { //... // These all have trivial implementation virtual void operator =(FibHeapNode& RHS); virtual int operator ==(FibHeapNode& RHS); virtual int operator <(FibHeapNode& RHS); }; class Event : public FibHeapNode { // These have nontrivial implementation virtual void operator=(FibHeapNode& RHS); virtual int operator==(FibHeapNode& RHS); virtual int operator<(FibHeapNode& RHS); }; class FibHeap { //... int DecreaseKey(FibHeapNode *theNode, FibHeapNode& NewKey) { FibHeapNode *theParent; // Some code if (theParent != NULL && *theNode < *theParent) { //... } //... return 1; } }; Much of FibHeap's implementation is similar: FibHeapNode pointers are dereferenced and then compared. Why does this code work? (or is it buggy?) I would think that the virtuals here would have no effect: since *theNode and *theParent aren't pointer or reference types, no dynamic dispatch occurs and FibHeapNode::operator< gets called no matter what's written in Event.

  Read the article

• C++ [] array operator with multiple arguments?

  - by genesys

  Can I define in C++ an array operator that takes multiple arguments? I tried it like this: const T& operator[](const int i, const int j, const int k) const{ return m_cells[k*m_resSqr+j*m_res+i]; } T& operator[](const int i, const int j, const int k){ return m_cells[k*m_resSqr+j*m_res+i]; } But I'm getting this error: error C2804 binary operator '[' has too many parameters

  Read the article

• Prolog: declaring an operator

  - by B K

  I have defined ! (factorial) function and registered it as arithmetic function and an operator, so that I can execute: A is 6!. Now I'd like to define !! (factorial of odd numbers), but the same way - writing clauses, registering arithmetic_function and operator, calling A is 7!! - results in SyntaxError: Operator expected How should I, if possible, register !! operator ? Yes, I realize, ! is normally the cut.

  Read the article

• operator[][] C++

  - by bobobobo

  I'd like to overload operator[][] to give internal access to a 2D array of char in C++. Right now I'm only overloading operator[], which goes something like class Object { char ** charMap ; char* operator[]( int row ) { return charMap[row] ; } } ; It works ok.. Is it possible to override operator[][] though?

  Read the article

• how to cout a vector of structs (that's a class member, using extraction operator)

  - by Julz

  hi, i'm trying to simply cout the elements of a vector using an overloaded extraction operator. the vector contians Point, which is just a struct containing two doubles. the vector is a private member of a class called Polygon, so heres my Point.h #ifndef POINT_H #define POINT_H #include <iostream> #include <string> #include <sstream> struct Point { double x; double y; //constructor Point() { x = 0.0; y = 0.0; } friend std::istream& operator >>(std::istream& stream, Point &p) { stream >> std::ws; stream >> p.x; stream >> p.y; return stream; } friend std::ostream& operator << (std::ostream& stream, Point &p) { stream << p.x << p.y; return stream; } }; #endif my Polygon.h #ifndef POLYGON_H #define POLYGON_H #include "Segment.h" #include <vector> class Polygon { //insertion operator needs work friend std::istream & operator >> (std::istream &inStream, Polygon &vertStr); // extraction operator friend std::ostream & operator << (std::ostream &outStream, const Polygon &vertStr); public: //Constructor Polygon(const std::vector<Point> &theVerts); //Default Constructor Polygon(); //Copy Constructor Polygon(const Polygon &polyCopy); //Accessor/Modifier methods inline std::vector<Point> getVector() const {return vertices;} //Return number of Vector elements inline int sizeOfVect() const {return vertices.size();} //add Point elements to vector inline void setVertices(const Point &theVerts){vertices.push_back (theVerts);} private: std::vector<Point> vertices; }; and Polygon.cc using namespace std; #include "Polygon.h" // Constructor Polygon::Polygon(const vector<Point> &theVerts) { vertices = theVerts; } //Default Constructor Polygon::Polygon(){} istream & operator >> (istream &inStream, Polygon::Polygon &vertStr) { inStream >> ws; inStream >> vertStr; return inStream; } // extraction operator ostream & operator << (ostream &outStream, const Polygon::Polygon &vertStr) { outStream << vertStr.vertices << endl; return outStream; } i figure my Point insertion/extraction is right, i can insert and cout using it and i figure i should be able to just...... cout << myPoly[i] << endl; in my driver? (in a loop) or even... cout << myPoly[0] << endl; without a loop? i've tried all sorts of myPoly.at[i]; myPoly.vertices[i]; etc etc also tried all veriations in my extraction function outStream << vertStr.vertices[i] << endl; within loops, etc etc. when i just create a... vector<Point> myVect; in my driver i can just... cout << myVect.at(i) << endl; no problems. tried to find an answer for days, really lost and not through lack of trying!!! thanks in advance for any help. please excuse my lack of comments and formatting also there's bits and pieces missing but i really just need an answer to this problem thanks again

  Read the article

• Why doesn't is operator take in consideration if the explicit operator is overriden when checking ty

  - by Galilyou

  Hey Guys, Consider this code sample: public class Human { public string Value { get; set;} } public class Car { public static explicit operator Human (Car c) { Human h = new Human(); h.Value = "Value from Car"; return h; } } public class Program { public static void Mani() { Car c = new Car(); Human h = (Human)c; Console.WriteLine("h.Value = {0}", h.Value); Console.WriteLine(c is Human); } } Up I provide a possibility of an explicit cast from Car to Human, though Car and Human hierarchically are not related! The above code simply means that "Car is convertible to human" However, if you run the snippet you will find the expression c is Human evaluates to false! I used to believe that the is operator is kinda expensive cause it attempts to do an actual cast that might result in an InvalidCastException. If the operator is trying to cast, then the cast should succeed as there's an operator logic that should perform the cast! What does "is" test? Does test a hierarchical "is-a" relationship? Does test whether a variable type is convertible to a type?

  Read the article

• operator<< cannot output std::endl -- Fix?

  - by dehmann

  The following code gives an error when it's supposed to output just std::endl: #include <iostream> #include <sstream> struct MyStream { std::ostream* out_; MyStream(std::ostream* out) : out_(out) {} std::ostream& operator<<(const std::string& s) { (*out_) << s; return *out_; } }; template<class OutputStream> struct Foo { OutputStream* out_; Foo(OutputStream* out) : out_(out) {} void test() { (*out_) << "OK" << std::endl; (*out_) << std::endl; // ERROR } }; int main(int argc, char** argv){ MyStream out(&std::cout); Foo<MyStream> foo(&out); foo.test(); return EXIT_SUCCESS; } The error is: stream1.cpp:19: error: no match for 'operator<<' in '*((Foo<MyStream>*)this)->Foo<MyStream>::out_ << std::endl' stream1.cpp:7: note: candidates are: std::ostream& MyStream::operator<<(const std::string&) So it can output a string (see line above the error), but not just the std::endl, presumably because std::endl is not a string, but the operator<< definition asks for a string. Templating the operator<< didn't help: template<class T> std::ostream& operator<<(const T& s) { ... } How can I make the code work? Thanks!

  Read the article

• Implementing Operator Overloading with Logarithms in C++

  - by Jacob Relkin

  Hello my friends, I'm having some issues with implementing a logarithm class with operator overloading in C++. My first goal is how I would implement the changeBase method, I've been having a tough time wrapping my head around it. My secoond goal is to be able to perform an operation where the left operand is a double and the right operand is a logarithm object. Here's a snippet of my log class: // coefficient: double // base: unsigned int // number: double class _log { double coefficient, number; unsigned int base; public: _log() { base = rand(); coefficient = rand(); number = rand(); } _log operator+ ( const double b ) const; _log operator* ( const double b ) const; _log operator- ( const double b ) const; _log operator/ ( const double b ) const; _log operator<< ( const _log &b ); double getValue() const; bool changeBase( unsigned int base ); }; You guys are awesome, thank you for your time.

  Read the article

• cast operator to base class within a thin wrapper derived class

  - by miked

  I have a derived class that's a very thin wrapper around a base class. Basically, I have a class that has two ways that it can be compared depending on how you interpret it so I created a new class that derives from the base class and only has new constructors (that just delegate to the base class) and a new operator==. What I'd like to do is overload the operator Base&() in the Derived class so in cases where I need to interpret it as the Base. For example: class Base { Base(stuff); Base(const Base& that); bool operator==(Base& rhs); //typical equality test }; class Derived : public Base { Derived(stuff) : Base(stuff) {}; Derived(const Base& that) : Base(that) {}; Derived(const Derived& that) : Base(that) {}; bool operator==(Derived& rhs); //special case equality test operator Base&() { return (Base&)*this; //Is this OK? It seems wrong to me. } }; If you want a simple example of what I'm trying to do, pretend I had a String class and String==String is the typical character by character comparison. But I created a new class CaseInsensitiveString that did a case insensitive compare on CaseInsensitiveString==CaseInsensitiveString but in all other cases just behaved like a String. it doesn't even have any new data members, just an overloaded operator==. (Please, don't tell me to use std::string, this is just an example!) Am I going about this right? Something seems fishy, but I can't put my finger on it.

  Read the article

• Visual C++ doesn't operator<< overload

  - by PierreBdR

  I have a vector class that I want to be able to input/output from a QTextStream object. The forward declaration of my vector class is: namespace util { template <size_t dim, typename T> class Vector; } I define the operator<< as: namespace util { template <size_t dim, typename T> QTextStream& operator<<(QTextStream& out, const util::Vector<dim,T>& vec) { ... } template <size_t dim, typename T> QTextStream& operator>>(QTextStream& in,util::Vector<dim,T>& vec) { .. } } However, if I ty to use these operators, Visual C++ returns this error: error C2678: binary '<<' : no operator found which takes a left-hand operand of type 'QTextStream' (or there is no acceptable conversion) A few things I tried: Originaly, the methods were defined as friends of the template, and it is working fine this way with g++. The methods have been moved outside the namespace util I changed the definition of the templates to fit what I found on various Visual C++ websites. The original friend declaration is: friend QTextStream& operator>>(QTextStream& ss, Vector& in) { ... } The "Visual C++ adapted" version is: friend QTextStream& operator>> <dim,T>(QTextStream& ss, Vector<dim,T>& in); with the function pre-declared before the class and implemented after. I checked the file is correctly included using: #pragma message ("Including vector header") And everything seems fine. Doesn anyone has any idea what might be wrong?

  Read the article

• What are the default return values for operator< and operator[] in C++ (Visual Studio 6)?

  - by DustOff

  I've inherited a large Visual Studio 6 C++ project that needs to be translated for VS2005. Some of the classes defined operator< and operator[], but don't specify return types in the declarations. VS6 allows this, but not VS2005. I am aware that the C standard specifies that the default return type for normal functions is int, and I assumed VS6 might have been following that, but would this apply to C++ operators as well? Or could VS6 figure out the return type on its own? For example, the code defines a custom string class like this: class String { char arr[16]; public: operator<(const String& other) { return something1 < something2; } operator[](int index) { return arr[index]; } }; Would VS6 have simply put the return types for both as int, or would it have been smart enough to figure out that operator[] should return a char and operator< should return a bool (and not convert both results to int all the time)? Of course I have to add return types to make this code VS2005 C++ compliant, but I want to make sure to specify the same type as before, as to not immediately change program behavior (we're going for compatibility at the moment; we'll standardize things later).

  Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >