Overloaded ostream operator in template

Question

I got confused in overloading the ostream operator<< for my template class. (unnecessary code deleted)

sparseArray2D.h:

#include <iostream> using namespace std; template <typename T> class sparseArray2D { private: //... public: //... friend ostream& operator << (ostream&, const sparseArray2D<T>&); //... } template <typename T> ostream& operator << (ostream& os, const sparseArray2D<T> &_matrix) { //... os<<"Overloaded operator works"; return os; };

and main:

#include "sparseArray2D.h" int _tmain(int argc, _TCHAR* argv[]) { //... sparseArray2D<int> *matrX = new sparseArray2D<int>(10, 9, 5); cout << matrX; //... }

No errors and no warnings in VS2012, but in the console I have 8 symbols as link or pointer at object. Like "0044FA80".

What's going wrong?

Note: A pointer is no object (C++)

user2249683
– user2249683

2015-04-10 18:36:46 +00:00
Commented Apr 10, 2015 at 18:36 — user2249683
– user2249683, Commented Apr 10, 2015 at 18:36
Why not sparseArray2D<int> matrX(10, 9, 5);

Neil Kirk
– Neil Kirk

2015-04-10 18:43:09 +00:00
Commented Apr 10, 2015 at 18:43 — Neil Kirk
– Neil Kirk, Commented Apr 10, 2015 at 18:43

Barry · Accepted Answer · 2015-04-10 18:36:44Z

2

That's because you're overloading (not reloading) on sparseArray2D<T>, but that is not what matrX is:

sparseArray2D<int> *matrX = new sparseArray2D<int>(10, 9, 5); // ^^ cout << matrX;

matrX is a pointer. As such, you're just streaming the pointer - which by default logs its address... which is apparently 0x0044FA80.

What you want is:

cout << *matrX;

answered Apr 10, 2015 at 18:36

Barry

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6 Comments

Archarious Over a year ago

Yes that is correct. Now I understand how it works. Thank you very much.

Qaz Over a year ago

Most likely, the OP doesn't really want a pointer at all. There's still the issue of the friend declaration mismatching the definition.

Archarious Over a year ago

Not absolutely understand what you mean

Qaz Over a year ago

@Archarious, To create a simple object, there's no need for a pointer at all (and Neil Kirk gave an alternative in another comment). Your definition of operator<< does not define the one you declared in your class, either, as evident by a linker error. I suggest reading about what that actually declares.

Archarious Over a year ago

@chris, ok, i understood! Thank you for explain and for article!

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