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### #1Burnt_Fyr  Members

Posted 21 June 2013 - 02:01 PM

What are the (dis)advantages to overloading an member operator vs a free operator? When and where is it best to use one or the other?

// Member functions

// In vector class
Vector3 operator * const (const Matrix&) {}

// in matrix class
Vector3 operator * const (const Vector&) {}

//vs

// free functions
Vector operator * (const Matrix&, const Vector&) {}
Vector operator * (const Vector& const Matrix&) {}



Posted 21 June 2013 - 02:32 PM

Prefer free functions unless they need to be members is the advice I have heard, you can always make them friend functions if you need to access private members. You can't do

builtin # classtype

with a member either (where builtin is a built-in type, # is the operator, and classtype is a class type), you can only do classtype # builtin.

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### #3__SKYe  Members

Posted 22 June 2013 - 02:46 AM

I think it is a matter of taste.

I, for one, prefer to use member operator overloading, because i find it less cumbersome, since, like Paradigm Shifter, the free operator must be a friend if you need to access private members, and  most of the times, i don't like having public members.

### #4Waterlimon  Members

Posted 22 June 2013 - 04:24 AM

I use class members because its less typing for me.
Then again, its kess typing because i put everything in headers because its less typing.

I did read some important c++ guru recommend making them free because it guarantees you they wont access private state or something.

o3o

### #5swiftcoder  Senior Moderators

Posted 22 June 2013 - 02:53 PM

What are the (dis)advantages to overloading an member operator vs a free operator? When and where is it best to use one or the other?

The exact same as for any other function:

- it should be a member function if it relies on private state.

- otherwise it should be a free function.

Tristam MacDonald - Software Engineer @ Amazon - [swiftcoding] [GitHub]

### #6Cornstalks  Members

Posted 22 June 2013 - 04:02 PM

Free functions, whenever possible (and practical), in my opinion. At least with that option, you can group all the functions into one header (if the left-hand and right-hand arguments are of different types); otherwise, you'll probably get one operator defined in one header and the other in another header. I prefer to keep my operators logically grouped together, which cannot always be done with member-function operators.

To expand on what ParadigmShifter mentioned, if you have operator* that multiplies a scalar (i.e. float, double, etc.) and a (math) vector, you'll probably have two operator* functions (so you can do (#1:) 3 * vector and (#2:) vector * 3). #1 cannot be a member function, but #2 can be. I like to group these two operators together, because the code for each of them is pretty much exactly the same (and they both do the exact same thing). The only way to group them together is to make both of them non-member functions. And for consistency's sake, that means I try to make all my operators non-member functions (or at least all that are practical to make non-member functions).

Some things, like operator=, cannot be a non-member function. Functions like these are, of course, implemented as member functions, because they legitimately need to be.

Edited by Cornstalks, 22 June 2013 - 05:36 PM.

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### #7SiCrane  Moderators

Posted 22 June 2013 - 07:52 PM

Note that the binary operators that have assignment versions, like *, - and +, rarely need to be implemented as member functions because they can be implemented in terms of the assignment versions. For instance, if you have *= then implementing operator * can be done as:
T operator*(const T & lhs, const T & rhs) {
T temp(lhs);
temp *= rhs;
return temp;
}


### #8Matt-D  Members

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