which "<<" operator function should be used?

Question

There are two ways of implementing operator "<<" or ">>" function in a project.

1.As a non-member function

2.As a friend

  #include<iostream>
  using namespace std;
  class xxx{
     private: int x;
     public: xxx(int val=0):x(val){}
              int getx(){return x;}
             friend ostream& operator <<(ostream& o, xxx& x1);
  };
  ostream& operator<<(ostream& o, xxx& x1)
  {
    o<<x1.getx();
    return o;
  }
  ostream& operator <<(ostream& o, xxx& x1)
  {
     o<<x1.getx();
     return o;
  }
  int main(int argc, char *argv[])
  {
     xxx x1(5);
     return 0;
  }

Looks like both non-member and friend function have same signature when implementing, and thus i get compiler error : "error: redefinition of 'std::ostream& operator<<(std::ostream&, xxx&)' ".

Could anyone please help how to compile the above code.

Also would like to know in which situation should we use non-member "operator =" function over friend "operator =" function.

If you declare a friend-function then it's really a non-member function as well. — Some programmer dude, Sep 26 '15 at 18:02
`operator =` can't be a non-member or friend; it has to be a member. — Alan Stokes, Sep 26 '15 at 18:13
@AlanStokes: if you make it as member function then you have to write like this x1< — Santosh Sahu, Sep 26 '15 at 18:33
possible duplicate of [Should operator<< be implemented as a friend or as a member function?](http://stackoverflow.com/questions/236801/should-operator-be-implemented-as-a-friend-or-as-a-member-function) — Csq, Sep 26 '15 at 18:37

Chris Dodd · Answer 1 · 2015-09-26T18:51:46.347

You seem to be confused -- a friend function is a non-member function. So your friend declaration declares the non-member function and makes it a friend. You then define the (non-member) function twice, which is an error.

The two ways to define (most) overloaded operators is as a member function or a non-member function. You cannot do both for the same operator, and if you define it as a non-member, it may be a friend or not as you prefer (friend is irrelevant)

As for your final question -- you cannot define operator= as a non-member function. It must be a member function. friend is irrelevant.

score 1 · Answer 2 · answered Sep 26 '15 at 18:11

1

The two definitions are identical. And in your case, the operator is not accessing private or protected members of the class, so the friend declaration is redundant.

answered Sep 26 '15 at 18:11

DrWatson

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which "<<" operator function should be used?

2 Answers2