Overloading Arithmetic Operators

Question

The assignment operator can be declared as

T& operator= (const t&);

in a class, but the arithmetic operators cannot be defined that way. It has to be fri

Answer 1

It is not mandatory that arithmetic operators should be friend

Well you can define like this:

MyClass MyClass::operator + (const MyClass& t) const
{
  MyClass ret(*this);
  ret += t;
  return ret;
}

The a + b is really a syntax sugar, the compiler will expand it to a.operator+(b). The previous sample will work if all your objects are MyClass instances, but will not work if you have to operate with others types, ie 1 + a, will not work, this can be solved by using friends.

MyClass operator + (int i, const MyClass& t)
{
  MyClass ret(i);
  ret += t;
  return ret;
}

This has to be done when the left hand side of the + operator is not a class, or it is a class but you can't add operator + to its definition.

Answer 2

I think that C++ FAQ Lite will give you a definitive answer.

Answer 3

They ideally should be globals and not necessarily friends, so that you can write:

yourtype v = 1;
yourtype w = 1 + v;

Since, 1 is not an object of yourtype, if the operator+ were a member it would throw a fit. However, making it global makes it convert the 1 to yourtype and then perform the operation. Making it a friend helps to extract and manipulate the members of yourtype as required -- though not required. As an example: You can implement the member function operator+= and use it in the implementation of the operator+.

Answer 4

The problem is that if you do something like this:

class A
{
    A& operator+(int n);
    // ...
}

int main()
{
    A my_var;
    int integer = 1;
    A + integer; // compiles
    integer + A // error: no function operator+(int, A) defined
}

it will not compile. A solution is to define operator+(int, A) and operator+(A, int) as friends of A. As a side note, the Boost Operators library makes this process very easy.