Is it not possible to call C++ operators manually?

I'm trying to understand operators in C++ more carefully.

I know that operators in C++ are b开发者_高级运维asically 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 operators [over.oper]

4. Operator functions are usually not called directly; instead they are invoked to evaluate the operators they implement (13.5.1 - 13.5.7). They can be explicitly called, however, using the operator-function-id as the name of the function in the function call syntax (5.2.2). [Example:

    complex z = a.operator+(b); // complex z = a+b;
    void* p = operator new(sizeof(int)*n);

—end example]

And operator functions require at least one parameter that is a class type or an enumeration type:

13.5 Overloaded operators [over.oper]

6. An operator function shall either be a non-static member function or be a non-member function and have at least one parameter whose type is a class, a reference to a class, an enumeration, or a reference to an enumeration.

That implies that an operator function operator+() that only takes ints cannot exist per 13.5/6. And you obviously can't use the function call syntax on an operator function that can't exist.

For basic types like int, float, double; the operators are already overloaded/pre-defined, so nothing special can be done for that. And,

int z = x + y;

is the only way to express/call it.

For interpretation purpose, actually both the statements,

int z = operator+(x,y);
int z = x.operator+(y);

are true (had it been overloadable).

Operator overloads only apply to objects and structs, not to fundamental types (such as int or float). If you had an object class like:

  class A {
    A operator+(const A& rhs) {
      return someComputedValue;
    }
  }

then you can indeed call myA.operator+(anotherA) and that will be equivalent to myA + anotherA.

You can't overload binary operators when both arguments are built in types. However for your own objects this is how you can create them.

//Simple struct that behaves like an int.
struct A
{
  int v_;
  explicit A(int i) : v_(i) {}  
  // A member operator so we can write a+b
  A operator+(const A & a ) const { return A( v_ + a.v_); }      
};

// A non-member operator, so we can write 1+a
A operator+(int i, const A & a)
{
   return A(i+a.v_);
}

int main()
{
  A a(1);
  A b(2);

  // Call the member version using its natural syntax    
  A c = a+b;
  //Call the member version using function call syntax
  A d = a.operator+(b);
  // Call the non-member version using the natural syntax
  A e = 1 + b;
  // Call the nonmember version using function call syntax.
  A f = ::operator+(1,b);
}

As has been mentioned in commentary and in other answers, by there is no operator+ for fundamental types. For classes, the answer to which of operator+(x,y) versus x.operator+(y) is correct is "it depends". Particularly, it depends on how operator+ was defined. If it was defined as an member function then you need to use x.operator+(y). If it was defined as a global function then you need to use operator+(x,y).

When the compiler confronts the statement z=x+y; your compiler is smart enough to look for the appropriate form. You shouldn't be expecting one or the other. You should be using x+y.

For native types, the operators aren't functions. Only overloaded operators are functions. Built-in operators are built-in - they don't have "functions", they usually just compile down to one or two assembly instructions that it would be insane to call as a function.

So neither operator+(x, y) nor x.operator+(y) is correct. I suppose x.operator+(y) is less correct because non-struct types can't have members, but I doubt that helps much.