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
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);
As Selman22 said, you are overriding the default object.Equals method, which accepts an object obj and not a safe compile time type. In order for that to happen, make your type implement IEquatable<Box>: public class Box : IEquatable<Box> { double height, length, breadth; public static bool operator ==(Box obj1, Box obj2) { if (ReferenceEquals(obj1, obj2)) … Read more
Overloadable Operators, from MSDN: Assignment operators cannot be overloaded, but +=, for example, is evaluated using +, which can be overloaded. Even more, none of assignment operators can be overloaded. I think this is because there will be an effect for the Garbage collection and memory management, which is a potential security hole in CLR … Read more
Have the assignment operator be callable on lvalues only: class C { // … public: C& operator=(const C&) & = default; }; Note the single ampersand after the closing parenthesis. It prevents assigning to rvalues.
I’d go with “C++ Coding Standards: 101 Rules, Guidelines, and Best Practices”: if you can do it as non-member function, do it as non-member function (in the same namespace). One of the reasons: it works better with implicit type conversion. An Example: You have a complex class with an overloaded operator*. If you want to … Read more
Not returning a reference is a waste of resources and a yields a weird design. Why do you want to do a copy for all users of your operator even if almost all of them will discard that value? a = b; // huh, why does this create an unnecessary copy? In addition, it would … Read more
Here’s a table of the Ruby operators. The ones that are methods and overloadable are: [ ] [ ]= Element reference, element set ** Exponentiation ! ~ + – Not, complement, unary plus and minus (method names for the last two are +@ and -@) * / % Multiply, divide, and modulo + – Plus … Read more
When both versions are available, the logic is pretty straightforward: const version is called for const objects, non-const version is called for non-const objects. That’s all. In your code sample a is a non-const object, meaning that the non-const version is called in all cases. The const version is never called in your sample. The … Read more
The technical reason why is found in ยง12.3.2: A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void. The rationale is (likely) … Read more