Passing pointer to function that accepts reference in c++

Question

I guess this is a n00b question because I couldn't find anything about it on web...

Here is Point class:

class Point {
public:
   Point();
   Point(double x, double y);


   double getX() const;
   double getY() const;
   void setX(double);
   void setY(double);

   friend std::ostream& operator<<(std::ostream& os, const Point& obj);
private:
   double x;
   double y;
};

And here is an implementation of operator<< function:

inline std::ostream& operator<<(std::ostream& os, const Point& obj) {
   os << "(" << obj.getX() << "," << obj.getY() << ")";
   return os;
}

Now, in main function I have Point *p;... How can I print it using std::cout?

Definitely a n00b question, but you shouldn't get penalized for it. — Mark Ransom, Sep 09 '14 at 16:04
well, I didn't mention that, but I have some experience with pointers so I tried this before and I got: `undefined reference to 'operator<<(std::ostream&, Point const&)'` — Sven Vidak, Sep 09 '14 at 16:49
By the way you don't need to make the streaming operator a friend, as it is implemented using only public getters. — CashCow, Sep 09 '14 at 18:24
Yeah, I know but NetBeans generated it for me with friend keyword so I left it... So, is there anything that can help me to solve this problem? I mean, I passed two courses where we were using C so I know how pointers work and this problem really confuses me. — Sven Vidak, Sep 09 '14 at 21:19

score 3 · Answer 1 · edited May 23 '17 at 12:21

3

You need to dereference your pointer but as pointers can be null, you should check first.

if( p != nullptr )
   std::cout << *p << std::endl;

or even just

if( p )
   std::cout << *p << std::endl;

And now, go and read this in our community wiki, hopefully it will provide you the answers.

What are the differences between a pointer variable and a reference variable in C++?

edited May 23 '17 at 12:21

Community

1
1

answered Sep 09 '14 at 16:03

CashCow

30,981
5
61
92

actually that link isn't in the community wiki but maybe it should be – CashCow Sep 09 '14 at 16:07
1

Do you mean to have `-` before `nullptr`? – Bathsheba Sep 09 '14 at 16:09
this works for me too, but I'm still getting undefined reference to `operator<<` function (using your code from main function) with my code – Sven Vidak Sep 10 '14 at 00:35
undefined reference is usually a linker error but you inlined it. Where in the code are all your files? – CashCow Sep 10 '14 at 09:49

score 0 · Accepted Answer · answered Sep 11 '14 at 22:29

So, I finally found out where was the problem.

Although all tutorials, books and even c++ reference agree that inline directive can be ignored by the compiler, it turns out that when I remove inline keyword from implementation of an overloaded function everything works.

Passing pointer to function that accepts reference in c++

2 Answers2