C++ cannot infer this automatically for a couple of reasons: It doesn’t make sense for every single type to be compared with operator<, so the type may not necessarily define a operator<. This means that operator== cannot be automatically defined in terms of operator< operator< isn’t required to compare its arguments. A programmer can define … Read more
Some operators return by value, some by reference. In general, an operator whose result is a new value (such as +, -, etc) must return the new value by value, and an operator whose result is an existing value, but modified (such as <<, >>, +=, -=, etc), should return a reference to the modified … Read more
The problem is that you declared operator>> and operator<< as non-member functions, but defined as a member function. This should fix that problem (but open another set of problems). So instead of ostream& Fraction::operator<<(ostream &os, Fraction& n) { … istream& Fraction::operator>>(istream &os, Fraction& n) { … implement as : ostream& operator<<(ostream &os, Fraction& n) { … Read more
Operator overloading is not a feature of Objective-C. If two instances of your classes can be added together, provide a method and allow them to be added using that method: Thing *result = [thingOne thingByAddingThing:thingTwo]; Or, if your class is mutable: [thingOne addThing:thingTwo];
Yes. Just override the object’s __iadd__ method, which takes the same parameters as add. You can find more information here.
(C++20 onward) With the acceptance of p1185 into C++20, you don’t need to provide more than one overload. The paper made these changes (among others) to the standard: [over.match.oper] 3.4 – […] For the != operator ([expr.eq]), the rewritten candidates include all member, non-member, and built-in candidates for the operator == for which the rewritten … Read more
They don’t have to be the same. Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completely up to the object in question what happens when one of these methods is called. If you use x += y but x does not provide an __iadd__ method (or … Read more
The postfix version of the increment operator takes a dummy int parameter in order to disambiguate: // prefix CSample& operator++() { // implement increment logic on this instance, return reference to it. return *this; } // postfix CSample operator++(int) { CSample tmp(*this); operator++(); // prefix-increment this instance return tmp; // return value before increment }
It sounds like you want to change: friend ostream& operator << (ostream& out, const Base<T>& e); To: template<class T> friend ostream& operator << (ostream& out, const Base<T>& e);
You get different behaviors because you are in fact calling two different (overloaded) sort functions. In the first case you call the two parameter std::sort, which uses operator< directly. Since the iterators to your vector elements produce non-const references, it can apply operator< just fine. In the second case, you are using the three parameter … Read more