Operation overloading for namespaced classes

Question

So let's say I have the following class :

namespace Example
{
    class Bar {};
}

Now in case I want to overload the operators for the class Bar, should I do :

 namespace Example
 {
    class Bar {};

    ostream& operator<<(ostream& os, const Bar& b)
    {/*..........*/}
}

or should I do :

namespace Example
{
    class Bar {};
}

ostream& operator<<(ostream& os, const Example::Bar& b)
{/*..........*/}

If I'm supposed to do either of the above, please post the explanation for why it should be done that way.

P.S. /*.....*/ simply means the body of the functions(omitted for simplicity)

Also possibly of interest: http://stackoverflow.com/questions/5195512/namespaces-and-operator-resolution — anderas, Jul 28 '15 at 13:17
@Petr IMO, no mine isn't a duplicate. In the link you gave it told me to do it in the namespace because the compiler would find it. But, the thing I"m asking is if either is better and why? Thank you for the link though — J.Alvaro.T, Jul 28 '15 at 13:18
@J.Alvaro.T, it also says "No need to pollute the global namespace". And other answers there say the same ("...and not to clutter the global namespace.") — Petr, Jul 28 '15 at 13:19
@Petr Thank you for pointing that out. Probably missed it. Sometimes I just skim through a question and missed the important parts. — J.Alvaro.T, Jul 28 '15 at 13:21

Bhavesh Kumar · Answer 1 · 2015-07-28T14:00:00.877

-2

  #include<iostream>
  using namespace std;

    namespace Example
     {
        class Bar
        {//........

friend ostream& operator<<(ostream& os, const Bar& b);
};


    }

    ostream& operator<<(ostream& os, const Example::Bar& b)
        {//..........
        }

    int main() {
        Example :: Bar b;
//        out<<b;  I am not sure About this 
        return 0;
    }

Hope I did well this time

edited Jul 28 '15 at 14:00

answered Jul 28 '15 at 13:18

Bhavesh Kumar

116
1
9

But that doesn't answer whether it should be placed outside or inside the namespace. – J.Alvaro.T Jul 28 '15 at 13:19
This does not answer the question. – Borgleader Jul 28 '15 at 13:19
ofcourse it answer as it replicate his second choice , he just mistaken to not referencing the definition of operator function with class Bar – Bhavesh Kumar Jul 28 '15 at 13:21
It doesn't answer the question, because the code wouldn't work at all. – juanchopanza Jul 28 '15 at 13:22
@juanchopanza , why ? – Bhavesh Kumar Jul 28 '15 at 13:23
Try it out. You'll see. – juanchopanza Jul 28 '15 at 13:31
@BhaveshKumar No it doesn't answer the question because the question is *why*. You just said "hey do this" without mentioning why this option is better than the other one. Also, you say "for convenience define outside class" well yeah that's already what he does. What OP wanted to know what should it be inside the namespace or outside of it, something you don't address at all. Which brings me back to: this does not answer the question. – Borgleader Jul 28 '15 at 13:35
@BhaveshKumar Try to use the operator. `Example::Bar b; std::cout << b;`. – juanchopanza Jul 28 '15 at 13:42
@juanchopanza ,Bogleader , please check my edited Answer , thanks for the knowledge guys . You corrected my Answer – Bhavesh Kumar Jul 28 '15 at 14:01

Operation overloading for namespaced classes

1 Answers1