Question

In C#, to overload an operator such as ‘+’, ‘-‘ etc, I have to make the function a static member of the class:

class MyType
{
   /*...*/

   public static MyType operator+ (MyType a, MyType b)
   {
       MyType ret;
       /* do something*/
       return ret;
   }
}

As far as I know, in C++ this is how I can overload an operator:

class MyType
{
   /*...*/

public:
   MyType operator+ (MyType b) // *this is the first operand
   {
       MyType ret;
       /* do something*/
       return ret;
   }
};

The problem is that *this is the first operand, so the first operand must be of type MyType. For example, if I want to add MyType to an integer:

MyType a, b;
b = a + 1;  // Valid
b = 1 + a;  // Error

In C#, I can overload the ‘+’ operator for each case.

My question is: can I do in C++ the same as in C#, use static operators? As far as I know, there is one way to do that, with friend operators, but they are lost when inheriting the function.

Answer 1

Make the operator+ overload with int on the left hand side a free function instead of a member function of MyType:

class MyType
{
  ...

  // MyType + int can be a member function because MyType
  // is the type of the sum's left hand side
  MyType operator+(int rhs) const;
};

// int + MyType needs to be a free function because
// int is the type of the sum's left hand side
MyType operator+(int lhs, const MyType &rhs);

Another common idiom is to make the overloads a friend of the class of interest. Now you can implement both cases in the same way:

class MyType
{
  ...

  friend MyType operator+(int lhs, const MyType &rhs)
  {
    // get access to MyType's private members here
    // to implement the sum operation
    ...
  }

  friend MyType operator+(const MyType &lhs, int rhs)
  {
    // you can also implement the symmetric case
    // of int on the right hand side here
    ...
  }
};

Note that even though the operator+ overloads look like member functions in the 2nd example, they are actually free functions that live in the global scope due to their declaration as friends of MyType.