Operator-overloading is defined as a function but not used as a function

Question

Consider the following C++ code such that I took it from "Sololearn". I'm a beginner and I would be grateful if any one could answer. It is on operator-overloading topic.

  class MyClass {
 public:
  int var;
  MyClass() {}
  MyClass(int a)
  : var(a) { }

  MyClass operator+(MyClass &obj) {
   MyClass res;
   res.var= this->var+obj.var;
   return res; 
  }
};

When they use it in main environment they use the following code

int main() {
  MyClass obj1(12), obj2(55);
  MyClass res = obj1+obj2;

  cout << res.var;
}

My questions:

Why does MyClass has two constructors? The most important question for me is that "operator+" has been defined like a function but when we use it in main its usage is not like a function. Why? Another different question is that is it true to say that the + used in this line MyClass res = obj1+obj2; is from the Operator+?

You can have as many constructors as you want, however it makes sense to construct an instance of your class. Yes that is the syntax to overload the `+` operator, not sure what else you want to hear. And yes in your example `MyClass::operator+` is being invoked from `obj1+obj2`. — Cory Kramer, Mar 24 '18 at 19:05
`MyClass res = obj1+obj2;` is the same as `MyClass res = obj1.operator+(obj2);` Overloaded operator are just functions with the addition of the "expected" synax. `operator+` being the name of the function. — Richard Critten, Mar 24 '18 at 19:08
`MyClass() {}` hello undefined behaviour. Don't write code like that kids! — n. m. could be an AI, Mar 24 '18 at 19:08
"Why does MyClass has two constructors?" - because you gave it two constructors: `MyClass() {} MyClass(int a)`.. what's unclear? — Jesper Juhl, Mar 24 '18 at 19:14
@JesperJuhl: The question is _why_ did the author give it two constructors. Notice that the OP is not that author. — Lightness Races in Orbit, Mar 24 '18 at 19:16

score 2 · Accepted Answer · answered Mar 24 '18 at 19:13

2

Why does MyClass has two constructors?

Because that's what the person who wrote it wanted. They wanted to make it so that you can provide an initial value if you wanted, and just let it default to zero otherwise.

Unfortunately, they made a mistake, and their default constructor is broken. It should be like this:

MyClass() : var(0) {}

…otherwise the value of var is indeterminate and a program trying to read it has undefined behaviour.

Now you can create an object like this:

MyClass obj;      // uses first constructor

Or like this:

MyClass obj(12);  // uses second constructor

"operator+" has been defined like a function but when we use it in main its usage is not like a function. Why?

Because you can. Because operators are special. C++ is designed this way. To be able to overload/define them they need to be functions, but to be operators they need their own syntax to use.

Note that you can call it like a function if you want to:

MyClass res = obj1.operator+(obj2);

…but you don't want to.

is it true to say that the + used in this line MyClass res = obj1+obj2; is from the Operator+?

Yes.

It would be better to learn C++ from a proper book.

answered Mar 24 '18 at 19:13

Lightness Races in Orbit

378,754
76
643
1,055

Thanks a lot for answering. Can we return another thing instead of res? Of course, I know the data type is MyClass. – Math Mar 25 '18 at 17:04
@Baghban: As long as it's a `MyClass` you can return whatever you like. Although people have certain expectations regarding how operator functions behave, and don't like to be surprised by them doing strange things. So try not to stray too far from what yours is doing at the moment. – Lightness Races in Orbit Mar 25 '18 at 18:46
I deleted the first constructor `MyClass(): var(0) {}` and got an error. Why? – Math Mar 26 '18 at 09:45
@Baghban: Because now the statement `MyClass res;` has no way to work. – Lightness Races in Orbit Mar 26 '18 at 09:54
(`MyClass res(this->var+obj.var);` probably would have been better) – Lightness Races in Orbit Mar 26 '18 at 09:55

Operator-overloading is defined as a function but not used as a function

1 Answers1