To get a value I would always use setter and getter. Would it be much better (performance) to use vector subscript operator overloading? And does vector subscription operator overloading require a size for the vector?
Recently, i successfully overloaded postfix operator to class counter by using Object and Class. Now, i want to overload same postfix operator to Inheritance. I created my code and generated in compiler. But, my compiler signaled me uncommon error(saw first time) and i couldn't generate any idea where my actually mistake is.
Here is my Code which objective is to count the number increasingly or decreasingly as per object created of CountDn class.
Code: #include <iostream> using namespace std; class Counter // base class { protected : // NOTE : Not Private unsigned int count;
[Code] ....
Error :|41|error: no 'operator++(int)' declared for postfix '++', trying prefix operator instead| |42|error: no 'operator++(int)' declared for postfix '++', trying prefix operator instead| |42|error: no match for 'operator=' in 'c2 = c1.CountDn::<anonymous>.Counter:perator++()'| |44|error: no 'operator--(int)' declared for postfix '--', trying prefix operator instead|
So i am having troubles with operator overloading in inherited class. Basically, it doesnt work. Consider this:
Code:
class A { public: A() { x=0; z= new int;
[Code] ....
Some how the copy constructor of a is improperly executed - the pointer is copied over, not re-created. As a result, the destructors crashes due to double-free.
*/ B bb = b; //doesnt work B bbb(b); //doesnt work B bbbb(b, 0); //works }
Above code shows the problem well. The "official" copy-constructor wont work - it copies over the pointer directly, and doesnt create a new one as it should. However, if i provide my own pseudo-copy-constructor that works. But ofcourse it's just a cheap work around - and wont actually work in real code (STL).
I am creating a class called time and we've had to do operator overloading for <, > , <=, >=, ==, !=, ++, --, >>, <<, * , +, and -.
Well I have done and error checked them all. The only one I cannot seem to get right is the minus and its because of the error checking. I am having issues with times like this
t1 = 0:0:2:3 t2 = 0:0:1:4
t1 - t2 should equal 0:0:0:59 but it returns 0:0:1:-1. (days:hours:minutes:seconds)
I need it to check for all cases and I just do not know how. Here is the code I have so far:
I wrote a simple Complex Class and overload input/output and +/- Operators in it!But there is something that doesn't work correctly!I can print an object of that class but I can't print sum of two object or something like that!
main3.cpp: In member function ‘FooB& FooB::operator=(const FooC&)’: main3.cpp:46:44: error: expected ‘(’ before ‘other’ main3.cpp:46:49: error: no matching function for call to ‘Foo<C>::Foo(const FooC&)’ main3.cpp:46:49: note: candidates are: main3.cpp:19:2: note: Foo<T>::Foo() [with T = C] main3.cpp:19:2: note: candidate expects 0 arguments, 1 provided main3.cpp:16:25: note: Foo<C>::Foo(const Foo<C>&) main3.cpp:16:25: note: no known conversion for argument 1 from ‘const FooC’ to ‘const Foo<C>&’
I designed a class template to create unique arrays. I was able to successfully input data to and output data from my array objects, irrespective of the datatype. However, I can't for the life of me fathom why my overloaded assignment operator worked perfectly well only for integer datatype and not for double/string datatypes.
Here is the class definition:
template <class dataType> class myArray { public: void setArrayData();
[code]....
And here is the definition of the overloaded assignment operator:
And here is my main function that tests the operations on objects of the class:
int main(){ //object declarations myArray<double> list(5); //a single-parameter object declaration of class myArray myArray<double> myList(2,13); //a two-parameter object declaration of class myArray
[code]....
The problem I'm having starts from where the assignment operator is being tested: for double and string datatypes, the upper input/output section works fine, but the assignment section freezes the display until the program execution is manually terminated!
I'm trying to get the command pattern for template classes down. I'm just having a hard time implementing one part of the code. I can't figure out why both the method being passed and the type accepted are not the same type.
SimpleCommand class template <class Receiver> class SimpleCommand : public Command { public: typedef void(Receiver::*Action)(); SimpleCommand(Receiver *r, Action a) :
[code].....
And their instantiation
MyClass *receiver = new MyClass; Command *aCommand = new SimpleCommand<MyClass>(receiver, &MyClass::Action); aCommand->Execute();
In that situation, << does not call the overloaded function, but rather calls the << method defined in the i/o library, which prints a message to the controlling terminal. So once it prints the message to the terminal, it then returns the out instance. Why return the out instance rather than a boolean like true? As you can see from the example, once the message is printed to terminal, out is not used anymore.
Code: #include <iostream>#include <iomanip> using namespace std; class Score { private: // Value at which we'll shift digits from million_counter to billion_counter static const int THRESHOLD = 1000000000;
[Code] ....
It gives the errors: line 105 error: million_counter was not declared in this scope line 106 error: normalizeScore was not declared in this scope line 110 error: million_counter was not declared in this scope and more of that until line 170 error: no match for 'operator<<' in 'std:perator<< <std::char_traits<char> >((* & std::cout), ((const char*)"a+b is ")) <<operator+((*c(const Score*) (& a)), (*(const Score*)(& b)))'
I thought that because i declared friend functions, they would be able to access the private variables of the class.
Well... I observed, as a non-professional programmer that "overloading operators" has some strict rules and some conventions... so any operator can differ from another. In order to have a clearest idea, I'd like to ask you to specify, for every operator, the correct (or best) way to overload it.
There are cases where you define &operator and cases where you define operator (without "&"). There are cases where operator are defined as "friend" inside class, and other cases where operator is declared externally.
example: ostream &operator<< (why it uses & ??)
So can we have a summary for all kind of operators?
I'm trying to overload operator<<, but I get an error saying 'ostream' does not name a type. Am I forgetting to declare something else? ostream& operator<< (ostream& out, Struct &b);I made sure to #include <iostream> too.
I am having a bit of an issue figuring out how to operator overload with chaining. I have this as my operator= function (Its for linked lists)
WORD & WORD::operator=(const WORD & Org){ cout << " operator= has been called WITH CHAINING "; character *p = front;
[Code] ....
I want to be able to do X = X = X where X is of class WORD, but it errors when that line is called. And by error, I dont mean a written error, it just compiles, then says 'MSVC has stopped working' on a new pop up.
How to finish these two remaining operator overloading functions
Also, "contents and NumItems are private"
Code: Bag operator+ (const Bag& b1, const Bag& b2); //Postcondition: the bag returned is the union of b1 and b2. ostream& operator<<(ostream&, const Bag&); //Overloading operator <<
I have a date class and i overloaded operator >> to accept input in dd/mm/yyyy format. if i enter the wrong date format my program will crash. How do i do exception handling for this? How should i do the try part? and for catch, I'll just catch a date class variable?
Code: void operator >> (istream &is, clsDate &date) { string inputDate; is >> inputDate; int mm = stringToNumber(inputDate.substr(3,2)); // read 2 characters from character number 3 start int dd = stringToNumber(inputDate.substr(0,2)); // read 2 characters from character number 0 start int yy = stringToNumber(inputDate.substr(6,4)); // read 4 characters from character number 6 start