C++ overloading `-` operator (as `* -1`)

Question

Say I have an int, like this:

int foo = 5;

I could then do this:

int bar = -foo;     // -5

I want to be able to do the same with my class, so how do I overload the - operator, used as * -1? Do I have to overload the * operator to do so?

I don't understand what the - and the * have to do with each other. — Joseph Mansfield
– Joseph Mansfield, Commented Apr 10, 2013 at 18:17
@JamesMcLaughlin Ah, that made me understand what the asker meant. — Joseph Mansfield
– Joseph Mansfield, Commented Apr 10, 2013 at 18:18
Something like the question here? stackoverflow.com/questions/2155275/… — Atrus
– Atrus, Commented Apr 10, 2013 at 18:20

john · Accepted Answer · 2013-04-10 18:18:08Z

4

class MyClass
{
    friend MyClass operator-(const MyClass& x);
};

or

class MyClass
{
    MyClass operator-() const;
};

Take your pick (although I would go for the first).

answered Apr 10, 2013 at 18:18

john

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5 Comments

MiJyn Over a year ago

Wow, that was simple! I'll accept it once I can verify it works.

Mike DeSimone Over a year ago

I'd go for the second, unless the - operator didn't need access to protected or private parts of MyClass, in which case I'd go for neither and just write MyClass operator-(const MyClass& x);. I prefer to avoid friend if at all possible.

john Over a year ago

@MikeDeSimone A global non-friend function is certainly the best if possible. I put friend there just to emphasize that this option does not involve a member function.

David Over a year ago

Why is a global non-friend function or friend function better than a member function? A member function means the declaration is in a place everyone is looking for functionality of the class. It's also logically a function of the class. What benefit is there?

john Over a year ago

@Dave There a real problem with member functions for symmetric binary operators (like +, *, < etc) in that the rules of C++ are such that different conversion rules apply to the left and right hand sides of your operator. This isn't a problem for operators which modify an existing object rather than return a new one (like +=). Unary - is somewhere in between, but since it creates a new object rather than modifies an existing one then stylistically I group it with binary +, - etc.

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