operator-keyword

This question already has an answer here: C extension: <? and >? operators 2 answers I came across the following code here , which is from the C++ implementation of Dijkstra algorithm using an adjacency matrix. //read in edges keeping only the minimum for(int i=0; i<E; i++) { int v1,v2,tmp; fin >> v1; fin >> v2; fin >> tmp; adjmat[v1][v2]<?=tmp; // <?= sets left to min(left,right) adjmat[v2][v1]<?=tmp; } Pay attention to the last two lines, which apply operator <?= . As being commented, the following line adjmat[v1][v2]<?=tmp; // <?= sets left to min(left,right) will set left to min(left,right

Is it possible to define a static insertion operator which operates on the static members of a class only? Something like: class MyClass { public: static std::string msg; static MyClass& operator<< (const std::string& token) { msg.append(token); return *this; // error, static } }; alternatively: static MyClass& operator<< (MyClass&, const std::string &token) { MyClass::msg.append(token); return ?; } This is how I would like to use it: MyClass << "message1" << "message2"; Thank you! If all the members of MyClass are static, it's possible to return a fresh instance. However, returning a

I'm trying to understand operators in C++ more carefully. I know that operators in C++ are basically just functions. What I don't get is, what does the function look like? Take for example: int x = 1; int y = 2; int z = x + y; How does the last line translate? Is it: 1. int z = operator+(x,y); or 2. int z = x.operator+(y); ? When I tried both of them, the compiler errors. Am I calling them wrong or are operators in C++ not allowed to be called directly? Using C++ standardese, the function call syntax ( operator+(x, y) or x.operator+(y) ) works only for operator functions : 13.5 Overloaded

Here is my code example: class X { public: void f() {} }; class Y : public X { public: X& operator->() { return *this; } void f() {} }; int main() { Y t; t.operator->().f(); // OK t->f(); // error C2819: type 'X' does not have an overloaded member 'operator ->' // error C2232: '->Y::f' : left operand has 'class' type, use '.' } Why the compiler is trying to "move the responsibility" for operator-> from Y to X? When I implement X::op-> then I cannot return X there - compile error says "infinite recursion" while returning some Z from X::op-> again says that Z doesn't have operator->, thus going

I hope someone might be able to answer why the following doesn't work. Bear with me though, I am still very much a noob... I just cannot get to the bottom of why the following using namespace std; #include <string> #include <iostream> class testClass { public: operator char* () {return (char*)"hi";}; operator int () {return 77;}; operator std::string () {return "hello";}; }; int main() { char* c; int i; std::string s = "goodday"; testClass t; c = t; i = t; s = t; cout<< "char: " << c << " int: " << i << " string: "<<s<<endl; return 0; } gives me a compile time error: myMain.cpp: In function

I have 3 dependent Rest API resources (lets say observables) like this: 1st observable produces one item as array of users, like this: getUsers(): Observable<User[]> [ { "id": 1, "name": "Peter", "surname": "Smith" }, { "id": 2, "name": "John", "surname": "Wayne" }, ... ] 2nd observable can be used to fetch addresses assigned to user, so the input parameter is User ID, and returns a one item as array of addresses: getUserAddresses(user_id: string): Observable<Address[]> [ { "id": 1, "city": "London", "street": "Wicombe 34" }, { "id": 2, "city": "Paris", "street": "La fever 20" }, ... ] 3rd