C++: Overloading overloaded operators

Aquarius_Girl · 01-04-2012, 03:14 AM

Code:

binaryOperators binaryOperators :: operator+ (const binaryOperators &right)
{
    return binaryOperators (*this + right.i);
}

binaryOperators operator+ (const binaryOperators &left, const binaryOperators &right)
{
    return binaryOperators (left.i + right.i);
}

int main ()
{
    binaryOperators obj (10);

    obj = 11 + obj;

    obj = obj + 11;

    return 0;
}

So, here the statement obj = 11 + obj; calls the function with explicit argument specification, and this one obj = obj + 11; calls the function which is the member of the class. Fine.

The problem is that second call results in an infinite loop.
What are the reasons and how to avoid that?

eSelix · 01-04-2012, 03:51 AM

Why

Code:

binaryOperators binaryOperators :: operator+ (const binaryOperators &right)
{
    return binaryOperators (*this + right.i);
}

and not

Code:

binaryOperators binaryOperators :: operator+ (const binaryOperators &right)
{
    return binaryOperators (this->i + right.i);
}

Aquarius_Girl · 01-04-2012, 03:53 AM

I get your point, but could you explain in detail why the previous call results in an infinite loop?

Proud · 01-04-2012, 06:26 AM

This.i not this? You're recursively calling object + thing's value, not object's value + thing's value?

johnsfine · 01-04-2012, 07:25 AM

Quote:

Originally Posted by Anisha Kaul

I get your point, but could you explain in detail why the previous call results in an infinite loop?

If you got that point, you wouldn't still be asking that question. *this + right.i is a recursive call to the function containing it (using an implicit call to the constructor to turn right.i back into an object of type binaryOperators).

In addition, your choice of two versions of operator+ doesn't make sense. The first is a non const member function, so it will cover all the cases where the left operand is a non const binaryOperators object. The second is not a member function but takes const binaryOperators objects for left operand.

Why would you want that second function and why would you want the first one to only take non const left operands?

The typical reason for a non member version of operator+ is to take a left operand type that is not a object of the class you are working on.

eSelix · 01-04-2012, 07:51 AM

I think that Anisha ask why compiler not warn about ambiguity between these two functions, when calling

Code:

binaryOperators + i_casted_for_binaryOperators

The first operator+ (class member) is something like

Code:

binaryOperators operator+ (binaryOperators left, const binaryOperators &right)

which probably for compiler is more appropriate then the second, and choose that version. But this is only my guess.

johnsfine · 01-04-2012, 08:16 AM

Quote:

Originally Posted by eSelix

I think that Anisha ask why compiler not warn about ambiguity between these two functions

I don't think Anisha was asking that. But since you are effectively asking it, I'll explain:

There is no ambiguity because one definition required a non const left operand and the other takes a const left operand.

When the left operand actually is const, the compiler can only use the definition that takes a const left operand. No hint of ambiguity there.

When the left operand is non const, a definition that takes a const left operand could be used (some ambiguity there) but the definition that takes a non const left operand is more specific. The ambiguity is resolved when one definition is more specific than the other.

Next consider the statement

Code:

    obj = 11 + obj;

The left operand is a const int and the right operand is a non const binaryOperators. No operator+ takes that pair of inputs, so the compiler must find implicit casts.

When you implicitly cast an integer to a binaryOperators, the result is a temporary object, and a temporary object passed by reference to a function is const. So even if the int had not been const, the binaryOperators object it becomes would be const.

The fact that the right operand is non const makes no difference because both operator+ definitions take only const right hand operands (so neither definition is a less specific fit to the right hand operand).

Only one definition fits the const left hand operand, so that definition is used.