Question

I found the URL below that says that

 If an operator can be used as either a unary or a binary 
 operator (&, *, +, and -), you can overload each use separately.

I am working with g++ in Linux and I tried the following and it didn’t compile.

int operator+ (const int a,const int b){
   std::cout << "MINE"<<std::endl;
   return 0;
}

int main(){
   char c='c';
   std::cout << c+2 << std::endl;
}

The error says

error: ‘int operator+(int, int)’ must have an argument 
of class or enumerated type

I was willing to play and see in action the Integer Promotion Rules.

Am I doing something wrong or that URL is valid only for MS or I misunderstood the promotion rule?

Answer 1

The error message indirectly tells you what you need to know — you are not permitted to overload operators (binary or unary) that act only on built-in types.

For a user-defined type T, you can separately overload binary + (for example by T operator+(T lhs, T rhs)) and unary + (for example by T operator+(T t)). You could also define operator+(T lhs, int rhs), but you can’t overload addition of two integers.