Is it possible to call the << operator using prefix notation?

Question

I am wondering if I could write, for instance: <<(object, cout); or <<(cout,object); where object is a user defined class which has the << operator overloaded, just as one could write: int a = +(2,3); and obtain the expected result, as in int a = 2 + 3;. Furthermore, what should one do if this is possible, but requires a few steps? Should one overload the << operator with two different signatures?

Answer 1

just as one could write: int a = +(2,3); and obtain the expected result, as in int a = 2 + 3;

No, you have a misunderstanding. +(2, 3) will go by the associativity of the comma operator and assign +3 (3) to the variable a.

Therefore, int a = +(2, 3) is the same as int a = 3, and not int a = 2 + 3

I am wondering if I could write, for instance: <<(object, cout); or <<(cout,object);

No, you can't.

If you want to use the operators in that fashion then you should call its fully qualified name, for example:

operator+ instead of +

operator<< instead of <<

Note:

This will not work for fundamental datatypes. Take care of the class scope and namespaces when you are using the overloaded versions of these operators.

Answer 2

You can do this, e.g.

operator<<(std::cout, "hi");

This expression

int a = +(2, 3);

is not doing operator+. It's first applying the , sequence operator, which gives 3, and then applying unary +, which gives 3.

You can't do this

int a = operator+(2, 3); // error

because ints are fundamental types.

If you have user defined types S, then the following snippets have the same meaning

S s{};
std::cout << s;
auto x = s + s;

is the same as

S s{};
operator<<(std::cout, s);
auto x = operator+(s, s);

assuming the operators are defined for S, the correct operator will be called.