C++ "<<" operator overload

Question

I've got this class:

// in Platform.h class Platform { private: float y; float xi; float xf; public: Platform(float y, float xi, float xf); virtual ~Platform(void); float getxi(); float getxf(); };

And I want to be able to do this:

Platform* p = new Platform(1.0,2.0,3.0); cout << p; // should it be *p?

I tried overloading the "<<" operator, like this:

// in Platform.cpp std::ostream& operator<<(std::ostream& out, Platform* p ) { out << "Platform: xi=" << p->getxi() << ", xf=" << p->getxf() << std::endl; return out; }

But this just prints a memory address (of course, because p is a pointer...). I'm quite sure that the above function isn't being called at all.

David Mokon Bond · Accepted Answer · 2012-12-13 04:06:03Z

3

Yes do a *p:

so cout << *p ;

And the conventional operator is...

std::ostream& operator << (std::ostream& out, const Platform& p) { }

You also need to export this in the header file so it is visible outside. Add:

friend std::ostream& operator<<(std::ostream& out, Platform* p ) ;

To Platform.h

edited Dec 13, 2012 at 4:06

answered Dec 13, 2012 at 3:51

David Mokon Bond

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5 Comments

Pubby Over a year ago

"Otherwise the template searches for and finds the pointer print function." No, his overload should be called instead. I'm guessing it wasn't visable from where he was calling from as Yakk said.

Maria Ines Parnisari Over a year ago

Should I add #include "Platform.cpp" in the file that calls cout << *p ?

David Mokon Bond Over a year ago

No, add a "friend std::ostream& operator<<(std::ostream& out, Platform* p ) ; " to Platform.hpp

Benjamin Bannier Over a year ago

@DavidMokonBond: Why would you want this to be a friend if it only uses public functions?

David Mokon Bond Over a year ago

Reasonable point, but I copied that from some of my code and I think its pretty standard for that having access to private data. Really depends though of course... If you have a nice system where you have gettters for everything you need sure...

Yakk - Adam Nevraumont · Accepted Answer · 2012-12-13 03:55:25Z

The overload declaration needs to be visible from where you use it. Change it to Platform const& and use *p while you are at it, because that is the conventional way.

fusha · Accepted Answer · 2012-12-13 03:58:18Z

Actually this works in g++ 4.7.2 but perhaps it depends on the compiler. You can do:

// in Platform.cpp std::ostream& operator<<(std::ostream& out, Platform& p ) { out << "Platform: xi=" << p.getxi() << ", xf=" << p.getxf() << std::endl; return out; }

And print using indirection (*p)

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