C++ Templates and a Linker Error - c++

Question

Please consider the following C++ program:
#include <iostream>
using namespace std;
template <class T> class Array
{
T *pType;
int itsSize;
public:
// template <class T>
friend ostream &operator<< (ostream &, Array<T> & );
};
template <class T>
ostream &operator<< (ostream &output, Array<T> &theArray)
{
return (output);
}
ostream &operator<< (ostream &, Array<int> &);
int main ()
{
Array<int> theArray;
cout << theArray << endl;
return 0;
}
The above code compiles but the linker gives the following error:
undefined symbol `operator<<(std::ostream&, Array&)'
I believe I need to tell the compiler to instantiate a function for the operator << but I do not know how to do that.
Bob

Answer 1

No, you don't need to tell the compiler to instantiate a template, it does that all by itself, if and when needed.
Your line
ostream &operator<< (ostream &, Array<int> &);
actually tells the compiler that you have some other operator for Array<int> so that it shouldn't use the template for that type.
So the linker goes looking for this other operator, but of course doesn't find any.
If you just remove that extra declaration, it should work much better.

Answer 2

The declaration
ostream &operator<< (ostream &, Array<int> &);
is not telling the compiler to instantiate the template. It is declaring a separate and distinct non-templated function i.e. overloading.
Within main(), the usage of operator<<() resolves to the declared function. The compiler does not need to specialize the template to find a match. However, since there is no definition (i.e. implementation) of that function, the linking fails.
Remove that declaration and you should find that your code compiles and links.
Incidentally, output streaming operators generally don't change the object being output. It is therefore advisable to change the templated ostream &operator<< (ostream &, Array<T> & ) to be ostream &operator<< (ostream &, const Array<T> & )
Also, since templates (typically) need to be defined in header files, and header files are included in multiple source files, it is better to avoid using namespace std, and replace ostream with std::ostream.

Answer 3

There are issues with your code, #BoPersson and #Perter pointed out some.
But even if you change what they suggested (Here is a live demo on ideone) and I ran it on my VS 2015 too. The compilers are still mad.
Ideone says:
prog.cpp:10:55: warning: friend declaration 'std::ostream& operator&)' declares a non-template function [-Wnon-template-friend]
friend ostream &operator & );
^
prog.cpp:10:55: note: (if this is not what you intended, make sure the function template has already been declared and add after the function name here)
/home/FHMhT9/ccNwcxP0.o: In function `main':
prog.cpp:(.text.startup+0x1b): undefined reference to `operator&)'
collect2: error: ld returned 1 exit status
And VS gives the pretty much same errors too.
Ideone is telling me, that the friend declaration in the class declares a non-template function.
So, change the declaration in the class to:
friend ostream &operator<< <T> (ostream &, Array<T> & );
// ^^^^^
// I added this <T> to tell the compiler that it is a template function.
But you need to define the function before this, so cut the implementation of operator<< and then paste it before the class. But you'd need a forward declaration of the class for that to work. So finally your code becomes pretty much like this:
And I changed the Array<T> & to const Array<T> & too.
#include <iostream>
template <typename T>
class Array;
template <class T>
std::ostream &operator<< (std::ostream &output, const Array<T> &theArray)
{
output << "abc";
return (output);
}
template <class T> class Array
{
T *pType;
int itsSize;
public:
// template <class T>
friend std::ostream &operator<< <T> (std::ostream &, const Array<T> &);
};
int main()
{
Array<int> theArray;
std::cout << theArray << std::endl;
return 0;
}
and this version now works.