Firstly, it should be noted that this has nothing to do with the operator being implemented as a friend specifically. It is really about implementing the copy-assignment as a member function or as a non-member (standalone) function. Whether that standalone function is going to be a friend or not is completely irrelevant: it might be, … Read more
Referential Equality Java In Java, the default implementation of equals compares the variable’s reference, which is what == always does: The equals method for class Object implements the most discriminating possible equivalence relation on objects; that is, for any non-null reference values x and y, this method returns true if and only if x and … Read more
Dart editor seems pretty happy with: operator [](int i) => list[i]; // get operator []=(int i, int value) => _list[i] = value; // set Try it in DartPad
Is it not supposed for a friend function to be explicitly defined outside of a class ? Friend functions can be defined (given a function body) inside class declarations. These functions are inline functions, and like member inline functions they behave as though they were defined immediately after all class members have been seen but … Read more
I think the best you’d be able to do is use IConvertible as a constraint and do something like: public static operator T +(T x, T y) where T: IConvertible { var type = typeof(T); if (type == typeof(String) || type == typeof(DateTime)) throw new ArgumentException(String.Format(“The type {0} is not supported”, type.FullName), “T”); try { … Read more
Why does this code compile? This compiles because the overridden interface method, Comparable<T>.compareTo, is itself an operator fun. /** * Compares this object with the specified object for order. Returns zero if this object is equal * to the specified [other] object, a negative number if it’s less than [other], or a positive number * … Read more
strict weak ordering This is a mathematical term to define a relationship between two objects. Its definition is: Two objects x and y are equivalent if both f(x, y) and f(y, x) are false. Note that an object is always (by the irreflexivity invariant) equivalent to itself. In terms of C++ this means if you … Read more
Should look like this: class Number { public: Number& operator++ () // prefix ++ { // Do work on this. (increment your object here) return *this; } // You want to make the ++ operator work like the standard operators // The simple way to do this is to implement postfix in terms of prefix. … Read more
I am just telling you about one other possibility: I like using friend definitions for that: namespace Math { class Matrix { public: […] friend std::ostream& operator<< (std::ostream& stream, const Matrix& matrix) { […] } }; } The function will be automatically targeted into the surrounding namespace Math (even though its definition appears within the … Read more
This is one of those frequently asked questions that have different approaches that are similar but not really the same. The three approaches differ in who you are declaring to be a friend of your function –and then on how you implement it. The extrovert Declare all instantiations of the template as friends. This is … Read more