How to define correctly unary operator- in C++

Question

Suppose we have this class definition:

class A{
public:
    int a;
    A(int i): a(i){}
    A operator-() const { return A(- a); }     # first operator- function
};
A operator-(const A& o) { return A(- o.a); }   # second operator- function

Now, in the main function, if I write:

A a1(10);
A a2 = -a1;                # Here the first operator- function (method) is called!

However, if I remove the first definition of operator- in the class, the second function is called. I would llike to know why the compiler prefers the first definition when both of them are provided.

I would like to know also why the compiler accept to call the first function (method). Because I think that the method is synonym to a1- and not -a1 : The object on which the method is applied comes before (at the left of) the operator.

You might want to read more about unqualified lookup, especially how it works with overloaded operators and overload resolution. — Some programmer dude
– Some programmer dude, Commented Jan 24, 2020 at 7:03

NutCracker · Accepted Answer · 2020-01-24 07:07:52Z

4

From the C++ standard, unary minus operator shall be defined like this:

T T::operator-() const;

inside class definition, and like this:

T operator-(const T &a);

outside class definition.

Now, you have both definition types in your class, so there member lookup comes in a play as specified here:

two separate lookups are performed: for the non-member operator overloads and for the member operator overloads (for the operators where both forms are permitted). Those sets are then merged with the built-in operator overloads on equal grounds as described in overload resolution. If explicit function call syntax is used, regular unqualified name lookup is performed

So, basically, member lookup finds A operator-() const;, non-member lookup finds A operator-(const A& o);. Overload resolution selects A operator-() const;.

answered Jan 24, 2020 at 7:07

NutCracker

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